Question

A man standing on the edge of the roof of a 20m tall building projects a ball of mass 100gm vertically up with the speed of 10m/s. The kinetic energy of the ball when it reaches the ground will be (Take ${\text{g = 10m/}}{{\text{s}}^{\text{2}}}$).
(A) 5${\text{J}}$
(B) 20${\text{J}}$
(C) 25${\text{J}}$
(D) zero

Answer 1

Hint: The ball is projected from height so it is projection from a height. Motion of a ball may be considered as the superposition of the two independent motions. Taking the equation of motion in vertical and horizontal direction will help in solving further.
Formula used: We will start by solving equations in vertical and horizontal direction.
We will also be using the equations of motion.
${\text{s = ut + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{g}}{{\text{t}}^{\text{2}}}$
${{\text{v}}^{\text{2}}}{\text{ = }}{{\text{u}}_{}}^{\text{2}}{\text{ + 2gs}}$

Complete step by step answer:
Here, we already know that there will be no change in magnitude of the horizontal component. Only the vertical component will change.
So, we start in the horizontal direction:
${\text{u = 10m/s}}$
Vertical motion with constant acceleration:
${\text{a = g = 10m/}}{{\text{s}}^2}$
The height of the cliff is considered to be ${\text{s = - 20}}$.
Similarly, we will find the vertical direction:
${{\text{v}}_{\text{y}}}^{\text{2}}{\text{ = }}{{\text{u}}_{\text{y}}}^{\text{2}}{\text{ + 2gs}}$
$ \Rightarrow {{\text{v}}_{\text{y}}}^{\text{2}} = {10^2} + 2 \times 10 \times - 20$
On solving, we get
$ \Rightarrow {{\text{v}}_{\text{y}}}^{\text{2}}{\text{ = 500}}$
Now, the kinetic energy,
 \[{\text{K}}{\text{.E = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{m}}{{\text{v}}^{\text{2}}}\]
 \[ \Rightarrow {\text{K}}{\text{.E = }}\dfrac{{\text{1}}}{{\text{2}}}{{ \times 1}}{{\text{0}}^{{\text{ - 1}}}}{{ \times 500 = 25J}}\]
Thus, the kinetic energy of the body is ${\text{25J}}$,

So, the correct option is C.

Note: The common mistake during the evaluation is doing the evaluation. It should be done carefully, as cases are different in case of ground-to-ground projection and projection from a height. Also, acceleration due to gravity $\left( {\text{g}} \right)$ always acts vertically downwards and there is no acceleration in horizontal direction unless mentioned. We need to take the value of $\left( {\text{g}} \right)$ to be $9.81m/{s^2}$ if not mentioned in the question. Generally, the value is considered to be $10m/{s^2}$ for the sake of calculation. Kinetic energy includes many forms of movement such as vibrations and rotations. Interesting Kinetic Energy Facts: An object keeps the same amount of kinetic energy unless it speeds up or slows down. Kinetic energy can be calculated for any moving object as long as the objects' mass and speed are known.