Ncert Exercises & Solutions

In a shotput event, an athlete throws the shotput of mass

10 kg

$10~\text{kg}$ with an initial speed of

1 m/s

$1~\text{m/s}$ at

45^{\circ}

$45^\circ$ from a height

1.5 m

$1.5~\text{m}$ above ground. Assuming air resistance to be negligible and acceleration due to gravity to be

10 {m/s}^{2}

$10~\text{m/s}^2$ , the kinetic energy of the shotput when it just reaches the ground will be:
1.

2.5 J

$2.5~\text{J}$
2.

5.0 J

$5.0~\text{J}$
3.

52.5 J

$52.5~\text{J}$
4.

155.0 J

$155.0~\text{J}$

Hint: Apply the concept of the work-energy theorem.

Given, $ h = 15 ~\text m~~,v = 1 ~\text{m/s},~~m= 10~\text{kg}, ~~g = 10 ~\text{m/s}^2 $

Step 1: Find the final kinetic energy.
From the conservation of mechanical energy.
$\left(\right. PE \left.\right)_ i + \left(\right. KE \left.\right)_ i = \left(\right. PE \left.\right)_ f + \left(\right. KE \left.\right)_ f$
$\Rightarrow mgh + \frac{1}{2} \left(mv\right)^{2} = 0 + \left(\right. KE \left.\right)_ f$
$\Rightarrow \left(\right. KE \left.\right) f = mgh + \frac{1}{2} \left(mv^{2}\right)$
$\Rightarrow \left(\right. KE \left.\right) _f = 10 \times 10 \times 1 .5 + \frac{1}{2} \times 10 \times \left( 1^{2} \right)$
$\Rightarrow 150 + 5 = 155 ~\text J$

Hence, option (4) is the correct answer. $\begin{array}{l} \end{array}$ $\begin{array}{l} \end{array}$

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