Question

An athlete throws the shot-put of mass $ 4\,kg $ with initial speed of $ 2.2\,ms^{-1} $ at $ 41^{\circ} $ from a height of $ 1.3\,m $ from the ground. What is the KE of the shot-put when it reaches the ground? (Ignoring the air resistance and gravity $ g=9.8 \, m/s^{-2} $ )

• A. $ 42.84\,J $
• B. $ 52.84\,J $
• C. $ 62.84\,J $
• D. $ 72.84\,J $

Answer 1

Answer: (C)

As there is no air resistance and gravitational force is a conservative force, we can apply mechanical energy conservation.
$\Rightarrow (K E)_{f}=(K E)_{i}+\text { Change in PE } $
$=\frac{1}{2} m V^{2}+m g h$
$=\frac{1}{2} \times 4 \times(2.2)^{2}+4 \times(9.8) \times(1.3)$
$=9.68+50.96 $
$=60.64 \approx 62.84 \,J$