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

Question

In a shotput event an athlete throws the shotput of mass 20 kg with an initial speed of 2 m s- 1 at 45o from a height 3m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m s- 2 , what will be the kinetic energy of the shotput when it just reaches the ground ? (A) 2.5 J (B) 5 J (C) 525 J (D) 640 J

Tardigrade Admin · Accepted Answer

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

$2.5 J$

$5 J$

$525 J$

$640 J$

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

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+Ki​⇒Kf​=640J

$2.5 J$

$5 J$

$525 J$

$640 J$

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