In a shotput event an athlete throws the shotput of mass 20 kg with an initial speed of 2 ms-1 at 45° from a height 3 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 ms -2 , the kinetic energy of the shotput when it just reaches the ground will be

Question

In a shotput event an athlete throws the shotput of mass 20 kg with an initial speed of 2 ms-1 at 45° from a height 3 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 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

Tardigrade Admin · Accepted Answer

Using law of energy conservation, (TME)i = (TME)f ⇒ Ui + Ki=Uf +Kf ⇒ 20 × 10 × 3 +(1/2) × 20 ×22 = 0 + Kf ⇒ Kf = 640 J

$2.5 J$

$5 J$

$525 J$

$640 J$

Using law of energy conservation,
$(TME)_{i} = (TME)_{f}$
$\Rightarrow U_{i} + K_{i} = U_{f} + K_{f}$
$\Rightarrow 20 \times 10 \times 3 + \frac{1}{2} \times 20 \times 2^{2}$
$= 0 + K_{f}$
$\Rightarrow K_{f} = 640 J$

2.5J

5J

525J

640J

Using law of energy conservation, (TME)i​=(TME)f​ ⇒Ui​+Ki​=Uf​+Kf​ ⇒20×10×3+21​×20×22 =0+Kf​ ⇒Kf​=640J

$2.5 J$

$5 J$

$525 J$

$640 J$

Using law of energy conservation,
$(TME)_{i} = (TME)_{f}$
$\Rightarrow U_{i} + K_{i} = U_{f} + K_{f}$
$\Rightarrow 20 \times 10 \times 3 + \frac{1}{2} \times 20 \times 2^{2}$
$= 0 + K_{f}$
$\Rightarrow K_{f} = 640 J$