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
- Zero
- 5 J
- 20 J
- 15 J
The Correct Option is B
Solution and Explanation
At the highest point, the vertical component of the velocity of the ball becomes zero, while the horizontal component remains the same. The initial kinetic energy at the point of projection is: \[ K_1 = \frac{1}{2} m v^2, \] where \( m = 0.1 \, \text{kg} \) and \( v = 20 \, \text{m/s} \). The initial velocity has two components:
- Horizontal component: \( v_x = v \cos \theta = 20 \cos 60^\circ = 10 \, \text{m/s} \),
- Vertical component: \( v_y = v \sin \theta = 20 \sin 60^\circ = 10\sqrt{3} \, \text{m/s} \). The kinetic energy at the highest point is: \[ K_2 = \frac{1}{2} m v_x^2 = \frac{1}{2} (0.1) (10)^2 = 5 \, \text{J}. \] Thus, the decrease in kinetic energy is: \[ \Delta K = K_1 - K_2 = \frac{1}{2} m v^2 - 5 = 20 - 5 = 5 \, \text{J}. \] Thus, the decrease in kinetic energy is \( \boxed{5 \, \text{J}} \).
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