Question

a football is thrown horizontally off the top of an apartment house that is 78.4 meters high. the ball is thrownat a speed of 15 m/s. how far will the ball travel

Answer 1

The motion of the ball is a composition of the two motions:
- on the horizontal axis, it is a uniform motion with constant speed
`v_x = 15 m/s`
- on the vertical axis, it is a uniformly accelerated motion with initial speed
`v_y =0` and constant acceleration
`g=9.81 m/s^2` toward the ground

The laws of motion on the two directions are:

`x(t) = v_x t`

`y(t) = h - (1)/(2)gt^2`
where h=78.4 m is the initial altitude of the ball.

We want to find how far the ball travels on the x-axis. In order to do that, we must find the time t at which the ball reaches the ground, i.e. the time t at which y(t)=0:

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

`t= \sqrt{ (2h)/(g) }= \sqrt{ (2 \cdot 78.4 m)/(9.81 m/s^2) }=4.0 s`

And so now we can find the distance x(t) covered by the ball during this time:

`x(t) = v_x t = (15 m/s)(4.0 s)=60 m`