Question

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

Answer 1

The total kinetic energy of the shotput when the shotput reaches the ground will be the sum of the kinetic energy given to the shotput and the potential energy of the shotput when it is thrown. And the air resistance is given as negligible, so the air resistance is left out.

Formula used:
The kinetic energy is given by,
$K.E = \dfrac{1}{2}m{v^2}$
Where $K.E$ is the kinetic energy of the shotput, $m$ is the mass of the shotput, and $v$ is the velocity of the shotput
The potential energy is given by,
$P.E = mgh$
Where $P.E$ is the potential energy of the shotput, $m$ is the mass of the shotput, $g$ is the acceleration due to gravity and $h$ is the height the shotput reaches.

Complete step by step answer:
Given that,
The mass of the shotput, $m = 10\,kg$
The velocity of the shotput, $v = 1\,m{s^{ - 1}}$
The angle of the shotput when it thrown, $\theta = {45^ \circ }$
The height of the shotput thrown, $h = 1.5\,m$
The acceleration due to gravity, $g = 10\,m{s^{ - 2}}$
Now, the total kinetic energy of the shotput is equal to the sum of the kinetic energy given to the shotput and the potential energy of the shotput. Then,
$E = K.E + P.E\,..................\left( 1 \right)$
Now substituting the formula of kinetic energy and the potential energy in the equation (1), then
$\Rightarrow E = \dfrac{1}{2}m{v^2} + mgh$
Now substituting the mass of the shotput, velocity of the shotput, acceleration due to gravity and the height of the shotput, then
$\Rightarrow E = \left( {\dfrac{1}{2} \times 10 \times {{\left( 1 \right)}^2}} \right) + \left( {10 \times 10 \times 1.5} \right)$
On simplifying the above equation, then
$\Rightarrow E = 5 + 150$
By adding the above equation, then
$\Rightarrow E = 155\,J$

The kinetic energy of the shotput when it just reaches the ground will be 155J. Hence, the option (D) is the correct answer.

Note:
When the shotput is thrown in the air, so it has some air resistance. But in the question, 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.