Question

A ball is thrown vertically upward at 10 m/s from the edge of a building that is 50 m high. How long does it take the ball to reach the ground?

Answer 1

It takes about 4.37 seconds.

Explanation:

To solve this we'll break the time down into two parts.

`t = 2t_1 + t_2`

with `t_1` being the time it takes the ball to go up from the edge of the tower and stop (it is doubled because it will take the same amount of time to return to 50m from the stopped position), and `t_2` being the time it takes the ball to reach the ground.

First we'll solve for `t_1`:

`10 - 9.8t_1 = 0`'
`9.8t_1 = 10`
`t_1 = 1.02` seconds

Then we'll solve for t_2 using the distance formula (note here that the velocity when the ball is heading down from the height of the tower is going to be 10 m/s toward the ground).
`d = vt_2 + 1/2at_2^2`
`50 = 10t_2 + 1/2*9.8t_2^2`
`0 = 4.9t_2^2 + 10t_2 - 50`

When solved, this polynomial equation yields either:
`t_2 = -4.37` seconds
or
`t_2 = 2.33` seconds

Only the positive one corresponds to a real physical possibility so we'll use that and solve.

`t = 2t_1 + t_2 = 2*1.02 + 2.33 = 4.37` seconds