Question

A cricket ball of mass $0.15kg$ is thrown vertically up by a bowling machine so that it rises to a maximum height of $20m$ after leaving the machine. If the part pushing the ball applies a constant force$F$ on the ball and moves horizontally a distance of $0.2m$ while launching the ball, the value of $F$ is ($g=10m{s}^{–2}$)

Answer 1

Step 1. Given data:

Mass of the ball, $m=0.15kg$

Maximum height, $h=20m$

Force applied by machine is $F$

Horizontal distance moved $x=0.2m$

acceleration due to gravity, $g=10m{s}^{–2}$

Step 2. Finding the initial velocity of the ball:

The velocity at maximum height is $0m/s$

Using$v^2-u^2=2as$ , we get

$v=\sqrt{2gh}=\sqrt{2\times 10\times 20}$

$v=20m/s$

Step 3. Finding the force:

The work done by machine is converted to Kinetic energy of the ball, equating the two,

$F.x=\frac{1}{2}m{v}^2$

$F\times 0.2=\frac{1}{2}\times 0.15\times {\left(20\right)}^2$

$F=150N$

Hence, the force applied by the machine is $F=150N$.