Question 1

An athlete participates in an interplanetary discus throw competition during an Olympiad that takes place on a planet where the free-fall acceleration is 9.7 m/s².He throws the discus with an initial velocity of 20 m/s at an angle of 60° from the vertical.Neglecting air resistance and the height of the discus at the point of release,what is the range of the discus? A)21 m B)60 m C)36 m D)40 m E)32 m

Accepted Answer

The range of the discus can be found using the equation:
R = (v^2/g) * sin(2θ)
where v is the initial velocity, g is the acceleration due to gravity (9.7 m/s^2 in this case), and θ is the angle of launch (60° in this case).

Plugging in the given values:
R = (20^2/9.7) * sin(2*60)
R = 36.28 m

Therefore, the range of the discus is approximately 36 m, which is closest to option C.

Question 2

A projectile is fired from ground level with a speed of 150 m/s at an angle 30° above the horizontal on an airless planet where g = 10.0 m/s².What is the horizontal component of its velocity after 4.0 s? A)150 m/s B)35 m/s C)130 m/s D)75 m/s E)38 m/s

Accepted Answer

First, we can find the initial horizontal and vertical components of the velocity:

Vx = 150 cos(30°) = 129.9 m/s
Vy = 150 sin(30°) = 75 m/s

Since there is no air resistance, the horizontal component of the velocity remains constant throughout the motion. Therefore, after 4.0 s, the horizontal velocity is still 129.9 m/s.

Alternatively, we can use the formula for horizontal displacement:

x = Vx * t = 129.9 m/s * 4.0 s = 519.6 m

Therefore, the horizontal component of the velocity after 4.0 s is 129.9 m/s or approximately 130 m/s (Choice C).

Question 3

At a location where g = 9.80 m/s²,a projectile returns to its original height after 4.08 seconds,during which time it travels 76.2 meters horizontally.If air can be neglected,what was the projectile's initial speed?

Accepted Answer

The answer of At a location where g = 9.80...

Question 4

A child throws a ball with an initial speed of 8.0 m/s at an angle of 40° above the horizontal.The ball leaves her hand 1.0 m above the ground.Neglect air (a)How far from where the child is standing does the ball hit the ground?
(b)How long is the ball in flight before it hits the ground?
(c)What is the magnitude of the ball's velocity just before it hits the ground?

Accepted Answer

The answer of A child throws a ball with an...

Question 5

A 5.0-kg stone is thrown upward at 7.5 m/s at an angle of 51° above the horizontal from the upper edge of a cliff,and it hits the ground 1.5 s later.Neglect air.Find the magnitude of its velocity vector just as it reaches the ground.

Accepted Answer

The answer of A 5.0-kg stone is thrown upward at...

Question 6

A child is trying to throw a ball over a fence.She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the horizontal.The ball leaves her hand 1.0 m above the ground and the fence is 2.0 m high.The ball just clears the fence while still traveling upwards..How far is the child from the fence if you neglect air?
A)0.73 m
B)1.6 m
C)2.7 m
D)3.8 m
E)7.5 m

Accepted Answer

Firstly, we need to find the maximum height the ball reaches. We can use the formula:
H = (V^2 sin^2θ) / 2g
where V is the initial velocity (8.0 m/s), θ is the angle above the horizontal (40°), and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we get:
H = (8.0^2 sin^2(40°)) / (2*9.8) = 1.91 m

Since the ball just clears the fence while still traveling upwards, we know that its height at that point is equal to the height of the fence (2.0 m). Therefore, we can use the same formula to find the horizontal distance traveled by the ball before reaching that height. This time, we want to find the distance traveled (x) instead of the maximum height, so we rearrange the formula:
x = (V^2 sin(2θ)) / g
Plugging in the values, we get:
x = (8.0^2 sin(2*40°)) / 9.8 = 1.6 m

Therefore, the child is 1.6 m from the fence. Hence, the answer is B.

Question 7

A rock is thrown from the roof of a building,with an initial velocity of 10 m/s at an angle of 30° above the horizontal.The rock is observed to strike the ground 43 m from the base of the building.What is the height of the building assuming no air?

Accepted Answer

The answer of A rock is thrown from the roof...

Question 8

A high-speed dart is shot from ground level with a speed of 150 m/s at an angle 30° above the horizontal.What is the vertical component of its velocity after 4.0 s if air effects can be neglected?
A)150 m/s
B)36 m/s
C)130 m/s
D)75 m/s
E)38 m/s

Accepted Answer

First, we need to find the initial vertical component of the velocity (vy). Using trigonometry, we have vy= v * sin(30°) = 150 m/s * 0.5 = 75 m/s.
Since air resistance is neglected, the vertical velocity of the dart does not change during its flight. Therefore, after 4 seconds, the vertical velocity remains the same and is equal to 75 m/s.

Question 9

A child throws a ball with an initial speed of 8.0 m/s at an angle of 40° above the horizontal.The ball leaves her hand 1.0 m above the ground.At what angle below the horizontal does the ball approach the ground assuming no effects of air?
A)35°
B)42°
C)48°
D)40°
E)65°

Accepted Answer

The vertical component of the initial velocity is:$$v_{y_i} = v_i \sin 	heta = (8.0 	ext{ m/s}) \sin 40° = 5.20 	ext{ m/s}$$The horizontal component of the initial velocity is:$$v_{x_i} = v_i \cos 	heta = (8.0 	ext{ m/s}) \cos 40° = 6.13 	ext{ m/s}$$The time it takes for the ball to reach the ground is:$$t = \frac{2v_{y_i}}{g} = \frac{2(5.20 	ext{ m/s})}{9.81 	ext{ m/s}^2} = 1.05 	ext{ s}$$The horizontal distance traveled by the ball is:$$x = v_{x_i}t = (6.13 	ext{ m/s})(1.05 	ext{ s}) = 6.44 	ext{ m}$$The angle below the horizontal at which the ball approaches the ground is:$$	heta = 	an^{-1} \left(\frac{v_{y_f}}{v_{x_f}}ight) = 	an^{-1} \left(\frac{-v_{y_i}}{v_{x_i}}ight) = 	an^{-1} \left(\frac{-5.20 	ext{ m/s}}{6.13 	ext{ m/s}}ight) = -42°$$The negative sign indicates that the ball is moving downward. Therefore, the angle below the horizontal at which the ball approaches the ground is 42°.

Question 10

An athlete competing in the long jump leaves the ground with a speed of 9.14 m/s at an angle of 35° above the horizontal.How long does the athlete stay in the air,assuming no effects of air?
A)0.50 s
B)0.88 s
C)1.1 s
D)2.5 s
E)0.54 s

Accepted Answer

The time of flight in projectile motion is given by the formula $T = \frac{2u\sin\theta}{g}$, where $u$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). Plugging in the given values: $T = \frac{2 \times 9.14 \times \sin(35^\circ)}{9.8}$, we find that the time of flight is approximately 1.1 seconds.

Question 11

An athlete competing in the long jump leaves the ground with a speed of 9.14 m/s at an angle of 55° with the vertical.What is the length of the athlete's jump if air is of no significance?
A)0.88 m
B)8.0 m
C)12 m
D)17 m
E)4.0 m

Accepted Answer

Using the formula for the horizontal displacement of a projectile: 
x = v₀ cosθ * t
We need to find the time it takes for the athlete to reach the ground again. 
Using the formula for the time of flight of a projectile: 
t = 2v₀ sinθ /g
Where g is the acceleration due to gravity (9.8 m/s²)
t = 2(9.14 m/s)sin55° / 9.8 m/s²≈1.38 s
Now we can find the horizontal displacement: 
x = (9.14 m/s)cos55° * 1.38 s ≈ 8.0 m

Question 12

The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively.If there is no air,how long does it take the football to reach the top of its trajectory?
A)1.0 s
B)2.0 s
C)3.0 s
D)4.0 s
E)5.0 s

Accepted Answer

The time taken by the football to reach the top of its trajectory can be found using the formula:
time = vertical component of initial velocity / acceleration due to gravity
= 20 m/s / 9.81 m/s^2
≈ 2.0 s (rounded to one decimal place)

Question 13

A projectile leaves the ground at 150 m/s and reaches a maximum height of 0.57 km.If there was no air,at what angle above the horizontal did it leave the ground?

Accepted Answer

The answer of A projectile leaves the ground at 150...

Question 14

A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal.Neglect air,take upward as the positive direction,and use g = 9.8 m/s². (a)Determine the time necessary for the projectile to reach its maximum height. (b)Determine the maximum height reached by the projectile. (c)Determine the horizontal and vertical components of the velocity vector at the maximum height. (d)Determine the horizontal and vertical components of the acceleration vector at the maximum height.

Accepted Answer

The answer of A projectile is fired from ground level...

Question 15

A wind farm generator uses a two-bladed propeller mounted on a pylon at a height of 20 m,as shown in the figure.The width of the pylon is very narrow,and the length of each propeller blade is 12 m.A tip of the propeller breaks off just when the propeller is vertical.The fragment flies off horizontally,falls,and strikes the ground at point P.Just before the fragment broke off,the propeller was turning uniformly,taking 1.2 s for each rotation.How far is point P from the base of the pylon if you neglect air? 
A)120 m
B)130 m
C)140 m
D)150 m
E)160 m

Accepted Answer

The answer of A wind farm generator uses a two-bladed...

Question 16

A boy kicks a football from ground level with an initial velocity of 20 m/s at an angle of 30° above the horizontal.What is the horizontal distance to the point where the football hits the ground if we neglect air?
A)20 m
B)35 m
C)18 m
D)60 m
E)30.0 m

Accepted Answer

The answer of A boy kicks a football from ground...

Question 17

A person throws a ball horizontally from the top of a building that is 24.0 m above the ground level.The ball lands 100 m down range from the base of the building.What was the initial velocity of the ball? Neglect air and use g = 9.81 m/s². A)202 m/s B)9.80 m/s C)19.6 m/s D)45.2 m/s E)94.4°

Accepted Answer

The answer of A person throws a ball horizontally from...

Question 18

A projectile is fired from ground level at an angle above the horizontal on an airless planet where g = 10.0 m/s².The initial x and y components of its velocity are 86.6 m/s and 50.0 m/s respectively.How long after firing does it take before the projectile hits the level ground? A)5.00 seconds B)10.0 seconds C)15.0 seconds D)20.0 seconds

Accepted Answer

First, we can use the y-component of the initial velocity to find the time of flight since the projectile will reach its maximum height at the midpoint of its motion and then return to the ground. Using the equation y = Vyi*t - 1/2 * g * t^2, where y = 0 (since the projectile returns to the ground), Vyi = 50.0 m/s and g = 10.0 m/s^2, we get t = 10.0 seconds.

Then, we can use the x-component of the initial velocity and the time of flight to find the horizontal distance traveled by the projectile before hitting the ground. Using the equation x = Vxi * t, where x is the distance traveled, Vxi = 86.6 m/s and t = 10.0 seconds, we get x = 866 m.

Therefore, the time taken for the projectile to hit the level ground is 10.0 seconds.

Question 19

A boy throws a ball with an initial velocity of 25 m/s at an angle of 30° above the horizontal.If you neglect air,how high above the projection point is the ball after 2.0 s?
A)5.4 m
B)13 m
C)25 m
D)43 m
E)50 m

Accepted Answer

The vertical component of the initial velocity is 25 m/s * sin(30°) = 12.5 m/s. After 2.0 s, the ball has traveled 12.5 m/s * 2.0 s = 25 m upward. The ball is also subject to gravity, which will cause it to fall 4.9 m/s^2 * (2.0 s)^2 / 2 = 9.8 m downward. Therefore, the ball is 25 m - 9.8 m = 15.2 m above the projection point after 2.0 s.

Question 20

A projectile is fired at an angle above the horizontal at a location where g = 9.8 m/s².The initial x and y components of its velocity are 86.6 m/s and 50 m/s respectively.At what angle was the projectile fired above the horizontal? A)45° B)60° C)30° D)90° E)75°

Accepted Answer

The angle $\theta$ of a projectile can be found using the formula $\tan(\theta) = \frac{V_y}{V_x}$, where $V_y$ and $V_x$ are the initial y and x components of the velocity, respectively. Substituting the given values, $\tan(\theta) = \frac{50}{86.6}$. Solving for $\theta$ gives $\theta \approx 30°$.

Quiz 4: Applying Newtons Laws

A projectile is fired from ground level with a speed of 150 m/s at an angle 30° above the horizontal on an airless planet where g = 10.0 m/s2.What is the horizontal component of its velocity after 4.0 s?

At a location where g = 9.80 m/s2,a projectile returns to its original height after 4.08 seconds,during which time it travels 76.2 meters horizontally.If air can be neglected,what was the projectile's initial speed?

A 5.0-kg stone is thrown upward at 7.5 m/s at an angle of 51° above the horizontal from the upper edge of a cliff,and it hits the ground 1.5 s later.Neglect air.Find the magnitude of its velocity vector just as it reaches the ground.

A rock is thrown from the roof of a building,with an initial velocity of 10 m/s at an angle of 30° above the horizontal.The rock is observed to strike the ground 43 m from the base of the building.What is the height of the building assuming no air?

A high-speed dart is shot from ground level with a speed of 150 m/s at an angle 30° above the horizontal.What is the vertical component of its velocity after 4.0 s if air effects can be neglected?

A child throws a ball with an initial speed of 8.0 m/s at an angle of 40° above the horizontal.The ball leaves her hand 1.0 m above the ground.At what angle below the horizontal does the ball approach the ground assuming no effects of air?

An athlete competing in the long jump leaves the ground with a speed of 9.14 m/s at an angle of 35° above the horizontal.How long does the athlete stay in the air,assuming no effects of air?

An athlete competing in the long jump leaves the ground with a speed of 9.14 m/s at an angle of 55° with the vertical.What is the length of the athlete's jump if air is of no significance?

The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively.If there is no air,how long does it take the football to reach the top of its trajectory?

A projectile leaves the ground at 150 m/s and reaches a maximum height of 0.57 km.If there was no air,at what angle above the horizontal did it leave the ground?

A boy kicks a football from ground level with an initial velocity of 20 m/s at an angle of 30° above the horizontal.What is the horizontal distance to the point where the football hits the ground if we neglect air?

A person throws a ball horizontally from the top of a building that is 24.0 m above the ground level.The ball lands 100 m down range from the base of the building.What was the initial velocity of the ball? Neglect air and use g = 9.81 m/s2.

A projectile is fired from ground level at an angle above the horizontal on an airless planet where g = 10.0 m/s2.The initial x and y components of its velocity are 86.6 m/s and 50.0 m/s respectively.How long after firing does it take before the projectile hits the level ground?

A boy throws a ball with an initial velocity of 25 m/s at an angle of 30° above the horizontal.If you neglect air,how high above the projection point is the ball after 2.0 s?

A projectile is fired at an angle above the horizontal at a location where g = 9.8 m/s2.The initial x and y components of its velocity are 86.6 m/s and 50 m/s respectively.At what angle was the projectile fired above the horizontal?

A projectile is fired from ground level with a speed of 150 m/s at an angle 30° above the horizontal on an airless planet where g = 10.0 m/s².What is the horizontal component of its velocity after 4.0 s?

At a location where g = 9.80 m/s²,a projectile returns to its original height after 4.08 seconds,during which time it travels 76.2 meters horizontally.If air can be neglected,what was the projectile's initial speed?

A person throws a ball horizontally from the top of a building that is 24.0 m above the ground level.The ball lands 100 m down range from the base of the building.What was the initial velocity of the ball? Neglect air and use g = 9.81 m/s².

A projectile is fired from ground level at an angle above the horizontal on an airless planet where g = 10.0 m/s².The initial x and y components of its velocity are 86.6 m/s and 50.0 m/s respectively.How long after firing does it take before the projectile hits the level ground?

A projectile is fired at an angle above the horizontal at a location where g = 9.8 m/s².The initial x and y components of its velocity are 86.6 m/s and 50 m/s respectively.At what angle was the projectile fired above the horizontal?