Question

A ball is thrown vertically upwards from a tower, strikes the ground after time t1. If it is thrown vertically downward with the same speed it strikes the ground after t2. If the ball is dropped from the top of the tower, after how much time it will strike the ground?

Answer 1

Answer:

√(t₁t₂)

Explanation:

In all the three cases displacement is same which is equal to the height of the tower.

Let the height of the tower is h

then

$h=-ut1+\frac{1}{2}gt1^{2}$             ............ (1)

$h=ut2+\frac{1}{2}gt2^{2}$             ............ (2)

From (1) and (2)

$h(\frac{1}{t1}+\frac{1}{t2})=\frac{1}{2} g$(t₁ + t₂)

or, h = (1/2)(gt₁t₂)                                               ............ (3)

again when the ball is dropped from the tower, let the time taken is t then

h = (1/2)gt²                                                   ............ (4)

From (3) and (4)

(1/2)gt² = (1/2)(gt₁t₂)    

or, t = √(t₁t₂)