A stone is thrown in a vertically upward direction with a velocity of 5 ms^-1. If the acceleration of the stone during its motion is 10 ms^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Given, initial velocity (u) = 5 ms^-1

Terminal velocity (v) = 0

Acceleration = -10 ms^-2 (as it is in upward direction)

According to the third equation of motion,

v² – u² = 2as

$\Rightarrow \text{s} = \dfrac{\text{v}^2 - \text{u}^2}{2\text{a}}$

Substituting we get,

s = $\dfrac{0 - {5}^2 }{2 \times -10} = \dfrac{-25}{-20}$ = 1.25 m.

Hence, distance covered = 1.25 m

According to the first equation of motion,

v = u + at

∴ time taken by the stone to reach a position of rest (maximum height) = $\dfrac{\text{v} - \text{u}}{\text{a}}$

Substituting we get,

$\dfrac{0-5}{-10}$ = 0.5 sec

∴ The stone reaches a maximum height of 1.25 m in 0.5 sec.