A stone is thrown in a vertically upward direction with a velocity of 5m/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 it take to reach there?

Explanation

Given initial velocity of stone, u= 5 m/s
Downward of negative acceleration `a= 10 m//s^(2)`
We know that, 2as `=v^(2) -u^(2)`
Therefore, Height attained by the stone
`2S = (0^(2) - 5^(2))/(-10) = (-5^(2))/(-10 xx 2)`
`S = (-25)/(-20) = 1.25m`
Also we know that final velocity, `v=u + " at or Time, " t = (V-u)/(a)`
Therefore, time, taken by stone to attain the height, `S = (0-5)/(-10) = 0.55s`