Question

A ball is thrown vertically upwards with an initial velocity of 20 m/s. If the acceleration due to gravity is \( 10 \, \text{m/s}^2 \), what is the maximum height reached by the ball?

• A. \( 10 \, \text{m} \)
• B. \( 20 \, \text{m} \)
• C. \( 40 \, \text{m} \)
• D. \( 80 \, \text{m} \)

Hint:

For vertical motion under gravity, the maximum height occurs when the velocity becomes zero. Use \( v^2 = u^2 + 2as \) or \( h = \frac{u^2}{2g} \) to calculate the height, ensuring consistent signs for acceleration.

Answer 1

The Correct Option is B

To find the maximum height reached by the ball, we use the kinematic equation: \[ v^2 = u^2 + 2as \] Where: - \( v = 0 \, \text{m/s} \) (final velocity at the maximum height, where the ball momentarily stops), - \( u = 20 \, \text{m/s} \) (initial velocity), - \( a = -10 \, \text{m/s}^2 \) (acceleration due to gravity, negative as it opposes the upward motion), - \( s \) is the maximum height (displacement). Substitute the values: \[ 0 = (20)^2 + 2 \times (-10) \times s \] \[ 0 = 400 - 20s \] \[ 20s = 400 \] \[ s = \frac{400}{20} = 20 \, \text{m} \] Alternatively, we can use the formula for maximum height in projectile motion: \[ h = \frac{u^2}{2g} \] \[ h = \frac{(20)^2}{2 \times 10} = \frac{400}{20} = 20 \, \text{m} \] Thus, the maximum height reached by the ball is \( 20 \, \text{m} \).