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 - (1)/(2) gt^2`
By requiring y(t)=0, we find the time t at which the ball reaches the floor:

`h- (1)/(2)gt^2=0`

`t= \sqrt{ (2h)/(g) }= \sqrt{ (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`