A ball of mass 100 g is projected with velocity 20 m/s at $60^\circ$ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is:
Show Hint
- 5 J
- 15 J
- 20 J
- zero
The Correct Option is B
Solution and Explanation
Concept:
In projectile motion:
The horizontal component of velocity remains constant.
The vertical component of velocity becomes zero at the highest point
Decrease in kinetic energy from projection to highest point is equal to the loss of kinetic energy associated with the vertical component of velocity.
Step 1: Given data: \[ m = 100\,\text{g} = 0.1\,\text{kg}, \quad u = 20\,\text{m/s}, \quad \theta = 60^\circ \]
Step 2: Vertical component of velocity: \[ u_y = u \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\,\text{m/s} \]
Step 3: Kinetic energy associated with vertical motion: \[ \Delta K = \frac{1}{2} m u_y^2 \] \[ \Delta K = \frac{1}{2} \times 0.1 \times (10\sqrt{3})^2 \] \[ \Delta K = 0.05 \times 300 = 15\,\text{J} \]
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