A ball with a horizontal speed of 0.91 m/s rolls off a bench 1.12 m above the floor. How long will it take the ball to hit the floor? Round to two decimal places and include units.

Answer 1

Solution:

The initial speed is V_o = 0.91 m/s and the initial height y₀ = 1.12 m. When the ball hit the floor the final position of the ball is y = 0 . The movement is horizontal, and therefore the angle of launching ∝ =0 . Using the equation of a projectile motion

y = -1/2gt² + V_o sin(∝)t + y₀ , and substituting 0 = -1/2 gt² + V₀ sin(0) t + 1.12m , and sin(0) =0, then the vertical component of the the elevation V₀ sin(0)t = 0

0 = -1/2gt² + 1.12 , solving for t , 0 = -1/2( 9.81 m/s²) t ² + 1.12 , 0 = -4.9 t² + 1.12, 4.9 t² = 1.12,

and t² = 1.12/4.9 = 0.229 s² , t = ±√0.229s² = ±0.478 s , neglect the negative answer because time cannot be negative, t = 0.478 rounded off to 2 decimal places = 0.48 s

Please note that the initial horizontal speed of the ball has no bearing on the time needed for the ball to reach the ground because the vertical component of the velocity vector is 0 . The time of fall in this case only depends on the gravitational acceleration and the initial height as in free fall