A projectile is fired horizontally with an initial speed of 50.0 m/s. Neglect air resistance.
-What is the magnitude of the displacement of the projectile 3.00 s after it is fired?
A)29.4 m
B)44.1 m
C)150 m
D)156 m
E)194 m

Question

Quizplus · Accepted Answer

The horizontal displacement of the projectile can be found using the formula: 
d = v*t 
where d is the displacement, v is the initial horizontal velocity (50.0 m/s), and t is the time (3.00 s). 
Thus, d = 50.0 m/s * 3.00 s = 150 m. 
However, we need to also find the vertical displacement of the projectile during this time. Since there is no air resistance, we know that the vertical velocity of the projectile will be affected only by the force of gravity. So we can use the formula: 
y = v0*t + (1/2)*a*t^2 
where y is the vertical displacement, v0 is the initial vertical velocity (0 m/s), a is the acceleration due to gravity (-9.81 m/s^2), and t is the time (3.00 s). 
Plugging in the values, we get: 
y = 0 m/s * 3.00 s + (1/2)*(-9.81 m/s^2)*(3.00 s)^2 = -44.1 m 
Note that the negative sign indicates that the displacement is downward. 
Therefore, the magnitude of the displacement can be found using the Pythagorean theorem: 
displacement = sqrt(d^2 + y^2) 
= sqrt((150 m)^2 + (-44.1 m)^2) 
= 156 m (rounded to three significant figures)

A Projectile Is Fired Horizontally with an Initial Speed of 50.0