Question

A body is projected vertically upwards. The times corresponding to height h while ascending and while descending are respectively , then, the velocity of projection will be (take g as acceleration due to gravity)

Answer 1

Answer:

velocity of projection is g(t₁ - t₂)/2

Explanation:

The complete question is :

A body is projected vertically upwards. The times corresponding to height h while ascending and while descending are t₁ and t₂ respectively , then, the velocity of projection will be (take g as acceleration due to gravity)?

Solution:

Let the initial velocity of projection is u

When the body is going up, using the second equation of motion

$h=ut_1-\frac{1}{2}gt_1^2$

Let the velocity at height h be v

Then using the first equation of motion

$v=u-gt_1$

when the body is descending, at height h its velocity will be v again

and when the body is descending downwards

$h=vt_2+\frac{1}{2} gt_2^2\\\implies h=\frac{1}{2} gt_2^2$

Therefore,

$ut_1-\frac{1}{2}gt_1^2=vt_2+\frac{1}{2}gt_2^2$

$\implies ut_1-\frac{1}{2}gt_1^2=(u-gt_1)t_2+\frac{1}{2}gt_2^2$

$\implies ut_1-\frac{1}{2}gt_1^2=ut_2-gt_1t_2+\frac{1}{2}gt_2^2$

$\implies ut_1-ut_2=\frac{1}{2}gt_1^2-gt_1t_2+\frac{1}{2}gt_2^2$

$\implies 2u(t_1-t_2)=g(t_1^2-2t_1t_2+t_2^2)$

$\implies 2u(t_1-t_2)=g(t_1-t_2)^2$

$\implies u=g(t_1-t_2)/2$