Question

A stone thrown vertically upwards with initial velocity u reaches a height 'h' before coming down. show that the time taken to go up is same as the time taken to come down.

Answer 1

When the stone is thrown from a tower say H metre high with an upward velocity u m/s, it will rise till the velocity of the stone reduces to zero under the action of acceleration due to gravity. The height gained is say h. This height can be calculated using the relation;

v² - u² = 2 g h. In this case v= 0, g = acceleration due to gravity taken as -g as it is in opposite direction to u. So h= u²/2g.

Let the height of the tower be H. The stone starts to drop from a height (H + h= H + u²/2g) to have a final velocity 3u on reaching the ground. Using the relation v² - intial velocity² = 2 g (H +h). In our case v= 3u, initial velocity with which the stone falls is zero. Substituting the values we get

9 u² -0²= 2 g (H + u²/2g)==> 2gH = 8u² or H = 4 u² /g .Let us check if the stone achieve a velocity 3u on reaching the ground. When the stone falls through a height h its velocity on reaching the tower top is u. So let us see the velocity of the stone in falling through H from top of the tower to the ground. At top of the tower the initial velocity of the stone is u, on reaching the bottom of the tower the velocity is v.

v² - u² = 2 g H == > v² = u² + 2 g × 4 u²/g = u² +8u²=9 u² , or v = ± 3 u . So indeed the stone arrives with a velocity 3u at the bottom of the tower.