Question

A stone is thrown vertically upward direction with a velocity of $5\,m{s^{ - 1}}$ if the acceleration of the stone during its motion is $10\,m{s^{ - 2}}$ in downward direction. What will be the height attained by the stone and how much time will it take to reach there?

Answer 1

Hint: In order to solve this question we need to understand the motion of stone thrown vertically upward. While throwing vertically upward only force of gravity acting downward direction hence acceleration of stone is in downward direction equal to acceleration due to gravity $9.8m{\sec ^{ - 2}}$ since velocity in upward direction is constant so acceleration in upward direction is zero. Also here we can use the equation of motion because acceleration is constant.

Complete step by step answer:
Let the velocity with which stone thrown upward be, $u = 5\,m{s^{ - 1}}$
Since the acceleration in downward direction is given as, $a = - 10\,m{s^{ - 2}}$
Here, negative sign signifies downward direction. Also we know that at highest point stone would be at rest so final velocity of the stone during upward motion is given as,
$v = 0\,m{s^{ - 1}}$
Let the stone cover distance, $S = h$
So using third equation of motion we get, ${v^2} = {u^2} + 2aS$
Putting values we get, $0 = ({5^2}) - (2 \times 10 \times h)$
$h = \dfrac{{25}}{{20}}m$
$\Rightarrow h = 1.25\,m$
So the maximum height it can cover is, $h = 1.25\,m$Let it takes “t” time to cover this distance then by using first equation of motion we get,
$v = u + at$
$\Rightarrow t = \dfrac{{v - u}}{a}$
Putting values we get, $t = \dfrac{{0 - 5}}{{ - 10}}s$
$\therefore t = 0.5\sec $

So the time required is $t = 0.5\sec $.

Note: It should be remembered that during motion in upward or downward direction we are neglecting air resistance for simpler calculation otherwise we know air resistance force is directly proportional to speed and surface area of ball on which it is acting and due to this the acceleration of body changes and we do not get the same result.