A particle is thrown up verticaly with a velocity with a velocity of `50 m//s`. what will be its velocity at the highest point of the journey ? How high would the particle rise ? What time would it take it take to reach the highest point ? Take `g=10m//s^(2)`.

Explanation

Here, intial velocity, `u=50m//s`, final velocity, `v=?`
height coverd, `h=?` time taken, `t=?`
`g=10m//s^(2)`
At the highest point, final velocity `v=0`.
From `v^(2)-u^(2)=2gh`, where `g=-10m//s^(2)` for upward journey,
`0-(50)^(2)=(-10)h`
or `h=-(2500)/(-20)=125m`
`v=u+g t, 0=50+(-10)t` or `t=50//10=5s`