an object thrown vertically upwards from the top of a building reaches the foot of the building in a time - Physics - Motion In A Straight Line

Question

an object thrown vertically upwards from the top of a building
reaches the foot of the building in a time t1 it takes a time t2,if
thrown vertically downwards with the same velocity if the time of
free fall is t show that t=under root ti*t2

Himanshu · Accepted Answer

Let h be the height of the tower $h = \frac{1}{2} g t^{2} f r e e f a l l$ $h = u t_{2} - \frac{1}{2} g t_{2}^{2} t h r o w n d o w n w a r d s$ $t_{1} = t_{2} + \frac{2 u}{g} \Rightarrow u = \frac{g}{2} (t_{1} - t_{2}) \frac{1}{2} g t^{2} = u t_{2} - \frac{1}{2} g t_{2}^{2} \Rightarrow \frac{1}{2} g t^{2} = \frac{g}{2} (t_{1} - t_{2}) t_{2} - \frac{1}{2} g t_{2}^{2} = \frac{g t_{1} t_{2}}{2} \Rightarrow t = \sqrt{t_{1} t_{2}}$

an object thrown vertically upwards from the top of a building reaches the foot of the building in a time t1.it takes a time t2,if thrown vertically downwards with the same velocity.if the time of free fall is t show that t=under root ti*t2.