A stone is thrown in a vertically upward direction with a velocity of `5 m//s`. If the acceleration of the stone during its motion is `10 m//s^(2)` in the downward direction, what will be the height attained by the stone and how much time will take to reach there ?

Explanation

Given initial velocity of stone, u = 5 m `s^(-1)`
Downward of negative Acceleration, a = 10 m `s^(2)`
We know that 2 as = `v^(2)-u^(2)`
Therefore, Height attained by the stone, `s=(0^(2))/(5^(2))xx(-10) m`
=`(-25)/(-20) m `
= 1.25 m
Also we know that final velocity, v = u + at
or, Time, `t=(v-u)/(a)`
Therefore, Time, t taken by stone to attain the height, s `(0-5)/(-10s)`
= 0.5 s