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?
A)35°
B)42°
C)48°
D)40°
E)65°

Question

Quizplus · Accepted Answer

First, we need to find the vertical component and horizontal component of the initial velocity:
Vertical component: 8.0sin40°≈5.1m/s
Horizontal component: 8.0cos40°≈6.1m/s

Using the vertical motion equation, we can find the time it takes for the ball to reach its highest point:
Δy = viy•t + 1/2•a•t²
0 = 5.1•t + 1/2•(-9.81)•t²
t≈0.5236s

Using the horizontal motion equation, we can find the total time of flight:
Δx = vix•t
Δx = 6.1•2•t ≈ 12.2t

Now we can use the vertical and horizontal components of velocity to find the final velocity:
vy = viy + a•t
vy = 5.1 + (-9.81)•0.5236 ≈ -0.65 m/s (negative because it is going downwards)
vx = vix
vf = √(vx² + vy²) ≈ 6.2 m/s

Finally, we can use trigonometry to find the angle below the horizontal:
tanθ = |vy|/vx
tanθ = 0.65/6.1
θ≈6.4°

However, this is the angle above the horizontal, so we need to subtract it from 40°:
40° - 6.4° ≈ 33.6°

Since the question asks for the angle below the horizontal, we subtract this value from 180°:
180° - 33.6° ≈ 146.4°

But we need to recognize that the ball is moving in the negative y-direction, so the angle below the horizontal is actually:
180° - 146.4° ≈ 33.6°

Therefore, the best choice is C) 48°.

A Child Throws a Ball with an Initial Speed of 8.0