Question

A ball thrown down from the top of a tower takes time t1 to reach the ground. It takes time t2 to reach
the ground if it is thrown from the same point with the same speed in the upward direction. Show that the
time it will take to fall freely to the ground from the top of the tower is
(t1t2)^1/2
. (Neglect air resistance)

Answer 1

Question 31703

1. First, let us consider the case when the initial velocity is directed downwards. Let $\nu_{0}$ denote the latter one. Hence, if vertical axis is directed upwards, the law of motion will be $h - \nu_{0}t_{1} - \frac{gt_{1}^{2}}{2} = 0$ (knowing that $t_{1}$ is given time).

2. Secondly, let us consider the case when the initial velocity is directed upwards. First, the body will move upwards until its velocity becomes zero. Let the time needed for this be $t'$ and the height above initial height be $h_1$ . Then, for full stop $v = v_0 - gt' = 0 \Rightarrow t' = \frac{v_0}{g}$ , and height $h_1$ is $h_1 = v_0 t' - g \frac{t'^2}{2} = \frac{v_0^2}{2g}$ (using previous formula for $t'$ ). After reaching maximum height ( $h + h_1$ ) and stop, body will move downwards with no initial velocity for some time $t''$ . The latter one is $t'' = \sqrt{2 \frac{(h + h_1)}{g}} = \sqrt{\frac{2}{g} \left( h + \frac{v_0^2}{2g} \right)}$ .

Full time of motion in case of initial velocity directed upwards is $t_2 = t' + t'' = \frac{v_0}{g} + \sqrt{\frac{2}{g} \left( h + \frac{v_0^2}{2g} \right)}$ .

In case of motion from height $h$ with no initial velocity, time is $t = \sqrt{2\frac{h}{g}}$ .

From points 1 and 2 one has following equations:

a) $h - v_{0}t_{1} - \frac{gt_{1}^{2}}{2} = 0$

b) $t_2 = t' + t'' = \frac{v_0}{g} + \sqrt{\frac{2}{g} \left( h + \frac{v_0^2}{2g} \right)}$

Let us solve b) for $\nu_{0}$ and plug it into a).

$$\frac {2}{g} \left(h + \frac {v _ {0} ^ {2}}{2 g}\right) = t _ {2} ^ {2} - 2 t _ {2} \frac {v _ {0}}{g} + \frac {v _ {0} ^ {2}}{g ^ {2}} \text{ from where } v _ {0} = \frac {g}{2 t _ {2}} \left(t _ {2} ^ {2} - \frac {2 h}{g}\right).$$

Now, plug latter formula for $\nu_{0}$ into a):

$$h - \frac {g}{2 t _ {2}} \left(t _ {2} ^ {2} - \frac {2 h}{g}\right) t _ {1} - \frac {g t _ {1} ^ {2}}{2} = 0, \text{ from here } h \left(\frac {t _ {1} + t _ {2}}{t _ {2}}\right) = \frac {g t _ {1} ^ {2}}{2} + \frac {g t _ {1} t _ {2}}{2} = \frac {g}{2} \left(t _ {1} ^ {2} + t _ {1} t _ {2}\right).$$

Hence, $h = \frac{g}{2} \left( \frac{t_2}{t_1 + t_2} \right) (t_1^2 + t_1 t_2) = \frac{g}{2} \frac{(t_2 t_1^2 + t_1 t_2^2)}{t_1 + t_2} = \frac{g}{2} \frac{t_1 t_2 (t_1 + t_2)}{t_1 + t_2} = \frac{g}{2} (t_1 t_2)$ .

Finally, knowing that $t = \sqrt{2\frac{h}{g}}$ , obtain $t = \sqrt{t_1t_2}$ .