Question

A ball of mass $0.2\text{kg}$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2\text{m}$ upward while applying the force and the ball goes upto $2\text{m}$ height further, find the magnitude of the force (in newtons).
[Take $g=10{\text{m/s}}^2$]

Answer 1

The situation is shown in the figure. At initial time, the ball is at $P$, then under the action of a force (exerted by hand), it moves from $P$ to $A$ and then from $A$ to $B$.

Let the acceleration of the ball during the part $PA$ be ${}^{\prime }{a}^{\prime }{\text{ms}}^{-2}$ (assumed to be constant) in upward direction and velocity of ball at $A$ be $v\text{ms}$.


Then, for $PA$, using equation of motion
$v^2=0^2+2a\times 0.2$

and for $AB$,
$0=v^2-2\times g\times 2$ [as ball comes to rest at $B$]
$\Rightarrow v^2=2g\times 2$
From the above equations,
$a=10g$


From FBD of the ball for $PA$
$F-mg=ma$
[$F$ is the force exerted by hand on ball]
$\Rightarrow F=m(g+a)$
$=0.2\left(11g\right)=22\text{N}$