Question

In a shot put event an athlete throws the shot put of mass 10 kg with an initial speed of $1m/s$ at ${45^ \circ }$ from a height $1.5m$ above ground. Assuming air resistance to be negligible and acceleration due to gravity to be $10m/{s^2}$ , the kinetic energy of the shot put when it just reaches the ground will be:
(A) $2.5J$
(B) $5.0J$
(C) $52.5J$
(D) $155.0J$

Answer 1

This question is solved by application of the law of conservation of energy. The air resistance is negligible and need not be subtracted from the final result.

Formula Used: The formulae used in the solution are given here.
$K.E = \dfrac{1}{2}m{v^2}$ where $K.E$ is the kinetic energy, $m$ is the mass of the shotput and $v$ is the velocity.
Potential energy of the shot-put of mass $m$ above the ground at a height $h$ is given by,
$P.E = mgh$ where $g$ is the acceleration due to gravity.

Complete Answer:
The law of conservation of energy states that energy can neither be created nor be destroyed; it can only be transformed from one form to another form.
In case of conservation of mechanical energy the potential energy (PE) of a body changes into kinetic energy (KE) and vice-versa , but the total mechanical energy (TME=PE+KE) remains constant.
It has been given that, in a shot-put event an athlete throws the shot-put of mass 10 kg with an initial speed of $1m/s$ at ${45^ \circ }$ from a height $1.5m$ above ground.
Thus, the shot-put possesses kinetic energy when in motion. The initial kinetic energy is given by,
$K.E = \dfrac{1}{2}m{v^2}$ where $K.E$ is the kinetic energy, $m$ is the mass of the shot-put and $v$ is the velocity.
Given that, $m = 10kg$ and $v = 1{m \mathord{\left/ {\vphantom {m s}} \right.} s}$ . Thus, kinetic energy initially is,
$K.{E_{initial}} = \dfrac{1}{2} \times 10 \times {\left( 1 \right)^2} = 5J$ .
Initial potential energy of the shot-put of mass $m$ above the ground at a height $h$ is given by,
$P.E = mgh$ where $g$ is the acceleration due to gravity.
Given that, $m = 10kg$ , $g = 10{m \mathord{\left/ {\vphantom {m {{s^2}}}} \right.} {{s^2}}}$ and $h = 1.5m$ . Thus, potential energy initially is,
$P.{E_{initial}} = 10 \times 10 \times 1.5 = 150J$ .
Since the total mechanical energy is given by the sum of the potential and kinetic energies of a body at a given time, the total initial energy of the shot-put is:
$K.{E_{initial}} + P.{E_{initial}} = 5J + 150J = 155J$ .
By law of conservation of energy, we can say that this initial energy is equal to the kinetic energy of the shot put when it just reaches the ground. At that time, height will be $h = 0m$ , thus potential energy is zero. The total energy is equal to the kinetic energy.
Thus, the kinetic energy of the shot put when it just reaches the ground is $155J$ .
So the correct answer is Option D.

Note:
This result is only applicable when air resistance is zero. It is given that the air resistance is negligible, so the air resistance is left out from the energy equation. If the air resistance is taken in the account, it should be subtracted from the total energy.