A ball is thrown horizontally from the roof of a building 8.6 m tall and lands 9.5 m from the base.

Question

What was the ball's initial speed?

Express your answer using two significant figures.

v0 = _________ m/s

Answer 1

Solution:

The initial height y_o = 8.6 m and the final distance X = 9.5 m. The ball is thrown horizontally meaning that the angle at which it was launched is zero, ∝ =0

Realizing that the horizontal component of the distance is X = V_o cos(∝) t = V_o cos ( 0) t = V_o t where V_o is the initial velocity. So the challenge here is to find t . Using the general equation of motion for the vertical component of the projectile

y = -1/2 gt² + V_o sin(∝)t + y_o and realizing that the final position of the ball when it is on the ground is

y = 0

Then 0 = -1/2 gt² + V_o sin(0) t + 8.6m , is reduced to 0 = -1/2 gt² + 8.6m since the y component of the force is zero, 0 = -1/2( 9.81 m/s²) t² + 8.6 , 0 = -4.9 t² + 8.6 , and 4.9 t² = 8.6 , t² = 8.6/4.9 = 1.755 s², and

t = ± 1.32 s. obviously the negative answer t = -1.32 s is neglected because time cannot be negative

and from the above equation X = V_o t , V_o = X / t = 9.5 m / 1.32 s = 7.19 m/s for two significant figures