Question

A particle is parojected vertically upwards from grund 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.

Answer 1

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)`.