Question

If a second ball were dropped from rest from height ymax, how long would it take to reach the ground

Answer 1

Answer:

$(b)\ t_1 - t_0$

$(d)\ t_2 - t_1$

$(e)\ \frac{t_2 - t_0}{2}$

Explanation:

Given

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Required

How long to reach the ground from the maximum height

First, calculate the time of flight (T)

$T =t_2 - t_0$

The time taken (t) from maximum height to the ground is:

$t = \frac{1}{2}T$

So, we have:

$t = \frac{t_2 - t_0}{2}$

Another representation is:

At ymax, the time is: t1

On the ground, the time is t2

The difference between these times is the time taken.

So;

$t = t_2 - t_1$

Since air resistance is to be ignored, then

$t_2 - t_1 = t_1 - t_0$ --- i.e. time to reach the maximum height from the ground equals time to reach the ground from the maximum height