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.

Explanation

1. Consider a stone thrown vertically upwards with initial velocity .u., reaches a maximum height .h.,
2. The time taken by a body thrown vertically upwards to reach the maximum height is called time of ascent. Let `t_(1)` be the time of ascent.
3. After that it changes its direction and falls freely to the earth.s surface. The stone is always under the influence of acceleration due to gravity .g. acting towards. Using Newton.s third equation of motion,
`V^(2)=u^(2)+2as`
Here, v = 0, a = -g, s = h
`therefore0=u^(2)+2(-g)h`
`thereforeh=(u^(2))/(2g)`............(i)
Also, v = u + at
Here, `0=u-"gt"_(1)`
`thereforet_(1)=(u)/(g)`............... (ii)
4. After reaching the maximum height, the body begins to travel downwards and reaches the ground.
5. The time taken by a freely falling body to touch the ground is called time to descent. Let `t_(2)` be the time of descent.
Here, Initial velocity (u) = 0, a = g and time `=t_(2)`
`s=ut+(1)/(2)at^(2)`
`thereforeh=(0)t_(2)+(1)/(2)"gt"_(2)^(2)`
`therefore(u^(2))/(2g)=(1)/(2)"gt"_(2)^(2)` [from ......(i)]
`thereforeu^(2)=g^(2)t_(2)^(2)`
TAking square root on both sides, we get, `u="gt"_(2)`
`therefore(u)/(g)=t_(2)`
`thereforet_(1)=t_(2)` [form ...... (ii) ]
Hence, time of ascent is equal to the time of descent.