Question

A ball of mass $0.2kg$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2m$ while applying the force and the ball goes up to 2 m in height further, find the magnitude of the force. Consider $g = 10m/{s^2}$
(A) 16 N
(B) 20 N
(C) 22 N
(D) 180 N

Answer 1

The final velocity attained by the ball as it leaves the hand is the initial velocity of the ascent to the maximum height. Take the hand to be at rest just before he tries to throw the ball.

Formula used: In this solution we will be using the following formula;
$H = \dfrac{{{u^2}}}{{2g}}$ where $H$ is the maximum height attained by an object from the point of release, $u$ is the initial velocity and $g$ is the acceleration due to gravity.
${v^2} = {u^2} + 2as$ where $v$ is the final velocity, $a$ is the acceleration of the object and $s$ is the distance moved to attain the final velocity

Complete step by step answer
The ball is said to be thrown vertically upward by hand which moved through a distance $0.2m$ before release. After the ball was released, it moved a further 2 m. hence the maximum height from the point of release is 2m. Thus, we can write
$H = \dfrac{{{u^2}}}{{2g}}$ where $H$ is the maximum height attained by an object from the point of release, $u$ is the initial velocity and $g$ is the acceleration due to gravity.
By inserting all known values we say
$2 = \dfrac{{{u^2}}}{{2\left( {10} \right)}}$
$\Rightarrow {u^2} = 40{m^2}/{s^2}$
This initial velocity is the final velocity of the hand movement. Then we can use the equation of motion
${v^2} = {u^2} + 2as$ where ${v^2} = {u^2} + 2as$ where $v$ is the final velocity, $a$ is the acceleration of the object and $s$ is the distance moved to attain the final velocity.
Hence, inserting all known values we get,
$40 = 0 + 2a\left( {0.2} \right)$ (since, in this case $v = u$ as calculated above) and assuming the hand was initially at rest.
Then, calculating for $a$
$a = \dfrac{{40}}{{2\left( {0.2} \right)}} = 100m/{s^2}$
Then the force required would be
$F = m(g + a)$
$\Rightarrow F = 0.2(10 + 100)$
By computation
$F = 22N$

Hence, the correct option is C.

Note
For clarity, the weight $mg$ was added to the total force because the total force applied must have to, at first, overcome the weight of the stone before being additionally able to accelerate. If the force was equal to the weight for an example, the motion of the hand would have moved at a constant velocity or not at all. Actually since the hand was at rest, then it would not have moved at all and the force would have been the normal force of the hand on the stone.