Question

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:

• A. Zero
• B. 5 J
• C. 20 J
• D. 15 J

Hint:

The decrease in kinetic energy in projectile motion corresponds to the loss of vertical velocity as the object reaches the highest point. The horizontal component remains unchanged.

Answer 1

The Correct Option is B

At its apex, the ball's vertical velocity is zero, while its horizontal velocity is constant. The initial kinetic energy at projection is \( K_1 = \frac{1}{2} m v^2 \), with \( m = 0.1 \, \text{kg} \) and \( v = 20 \, \text{m/s} \). The initial velocity components are: - Horizontal: \( v_x = v \cos \theta = 20 \cos 60^\circ = 10 \, \text{m/s} \); - Vertical: \( v_y = v \sin \theta = 20 \sin 60^\circ = 10\sqrt{3} \, \text{m/s} \). The kinetic energy at the highest point is \( K_2 = \frac{1}{2} m v_x^2 = \frac{1}{2} (0.1) (10)^2 = 5 \, \text{J} \). Consequently, the reduction in kinetic energy is \( \Delta K = K_1 - K_2 = \frac{1}{2} m v^2 - 5 = 20 - 5 = 5 \, \text{J} \). The decrease in kinetic energy is \( \boxed{5 \, \text{J}} \).