Ncert Exercises & Solutions

In a shotput event, an athlete throws the shotput of mass $10$ kg with an initial speed of $1$ m/s at $45^\circ$ from a height $1.5$ m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be $10$ m/ $s^{2}$ , the kinetic energy of the shotput when it just reaches the ground will be:
1. $2.5$ J
2. $5.0$ J
3. $52.5$ J
4. $155.0$ J

(d) Hint: Apply the concept of work-energy theorem.

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

Step 1: Find the final kinetic energy.

From the conservation of mechanical energy.

$\begin{array}{l} (PE) i + (KE) i = (PE) f + (KE) f \\ \Rightarrow mgh + \frac{1}{2} {mv}^{2} = 0 + (KE) f \\ \Rightarrow (KE) f = mgh + \frac{1}{2} {mv}^{2} \\ \Rightarrow (KE) f = 10 \times 10 \times 1 .5 + \frac{1}{2} \times 10 \times (1)^{2} \\ = 150 + 5 = 155 J \end{array}$

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