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)

$62.84\,J$

Answer 2

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