A particle is projected vertically upwards from ground with initial velocity `u`.
a. Find the maximum height `H` the particle will attain and time `T` that it will attain and time `T` that it will take to return to the ground .
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b. What is the velocity when the particle returns to the ground?
c. What is the displacement and distance travelled by the particle during this time of whole motion.

Explanation

Cosnider the motion from `A` to `B`:
`s=+H` (final point lies above the initial point), initial velocity `=u`, final velocity `v=0`.
Let the time taken to go from `A` to `B` be `t_(1)`.
Using `v=u-g t`, we get
`0=u-g t_(1) rArrt_(1)=(u)/(g)`
Using `s=ut-(1)/(2)g t^(2)`,
`rArr h=u ((u)/(g))-(1)/(2)g((u)/(g))^(2)=(u^(2))/(2g)`
So the maximum height attained is `H=(u^(2))/(2g)`.
Cosnsider the return motion from `B` to `A`:
`s=H` (final point lies below the initial point) `u=0` (at point `B`, velocty is zero)
Let time taken to go from `B` to `A`be `t_(2)`. We have
`t_(2)=sqrt((2H)/(g))=sqrt((2)/(g)(u^(2))/(2g))(u)/(g)`
Hence `t_(1)` is known as the time of ascent and `t_(2)` is known as the time of descent. We can see that time of ascent=Time of descent`=(u)/(g)`
Total time flight `T=t_(1)+t_(2)=(2u)/(g)`
Hence, time of flight is the time for which the particle remains in air.
Alternative method to find the time of flight:
a. Consider the motion from `A` to `B`:
`s=0` (initial and final points are same )
Initial velocity `=u`, time taken`=T`
Using `s=ut-(1)/(2)g t^(2)`, we have `0=uT-(1)/(2)g t^(2)`
`rArr T =(2u)/(g)`
b. Magnitude of velocity on returning the fround will be same as that of inital velocity but directionwill be opposite.
Proof:Let `v` be the velocity on reaching the ground. Then from previous formulae, we get
` v=- sqrt(2gH)=-sqrt(2g(u^(2))/(2g))rArrv=-u`, hence proved,
c. Displacement`=0`, distance travelled `=2H=u^(2)/(g)`.