Question

In a shotput event an athlete throws the shotput of mass 20 kg with an initial speed of $45 ^ { \circ }$from a height 3 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be $10 \mathrm {~ms} ^ { - 2 }$, the kinetic energy of the shotput when it just reaches the ground will be:

• A. 2.5 J
• B. 5 J
• C. 525 J
• D. 640 J

Answer 1

Answer: (D)

640 J

Answer 2

The situation is as shown in the figure.

Maximum height reached by the shotput is

Here, u$\mathrm { u } = 2 \mathrm {~ms} ^ { - 1 } , \theta = 45 ^ { 0 }$

$\therefore \quad \mathrm { H } _ { \max } = \frac { ( 2 ) \sin ^ { 2 } 45 ^ { 0 } } { 2 \times 10 } = \frac { 4 } { 40 } = 0.1 \mathrm {~m}$

Potential energy at

H = 3g (3 + 0.1)

$= ( 20 \mathrm {~kg} ) \left( 10 \mathrm {~ms} ^ { - 2 } \right) 3.1 \mathrm {~m} = 620 \mathrm {~J}$

Kinetic energy at

$= \frac { 1 } { 2 } \times 20 \times \left( 2 \times \frac { 1 } { \sqrt { 2 } } \right) ^ { 2 } = \frac { 40 } { 2 } = 20 \mathrm {~J}$

Total mechanical energy at H = 620 J + 20 J = 640 J

Let $\mathrm { K } _ { \mathrm { G } }$ is the kinetic energy at G.

Potential energy at G = 0

Total mechanical energy at $\mathrm { G } = \mathrm { K } _ { \mathrm { G } } + 0$

By the law of conservation of mechanical energy between H and G is

$620 \mathrm {~J} + 20 \mathrm {~J} = \mathrm { K } _ { \mathrm { G } } + 0$

$\therefore \mathrm { K } _ { \mathrm { G } } = 640 \mathrm {~J}$