Question

A ball with a horizontal speed of 1.0m/s rolls off a bench 2.0 m high. (a) how long will the ball take to reach the floor? (b) how far from a point on the floor directly below the edge of the bench will the ball land?

Answer 1

The motion of the ball is a composition of two motions:
- on the x (horizontal) axis, it is a uniform motion with initial velocity $v_x = 1.0 m/s$
- on the y (vertical) axis, it is a uniformly accelerated motion with acceleration $g= 9.81 m/s^2$ 

(a) to solve this part, we just analyze the motion on the vertical axis. The law of motion here is
$y(t) = h - \frac{1}{2} gt^2$
By requiring y(t)=0, we find the time t at which the ball reaches the floor:
$h- \frac{1}{2}gt^2=0$
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2\cdot 2.0 m}{9.81 m/s^2} }=0.64 s$

(b) for this part, we can analyze only the motion on the horizontal axis. To find how far the ball will land, we must calculate the distance covered on the x-axis, x(t), when the ball reaches the ground (so, after a time t=0.64 s):
$x(t) = v_x t = (1.0 m/s)(0.64 s)=0.64 m$