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

The maximum height attained by the ball is determined using the kinematic equation:\[v^2 = u^2 + 2as\]Key variables are defined as follows: \( v = 0 \, \text{m/s} \) (final velocity at peak altitude), \( u = 20 \, \text{m/s} \) (initial velocity), \( a = -10 \, \text{m/s}^2 \) (acceleration due to gravity, acting downwards), and \( s \) (maximum height). Substituting these values yields:\[0 = (20)^2 + 2 \times (-10) \times s\]\[0 = 400 - 20s\]\[20s = 400\]\[s = \frac{400}{20} = 20 \, \text{m}\]An alternative method involves the direct formula for maximum height in projectile motion:\[h = \frac{u^2}{2g}\]Calculation using this formula:\[h = \frac{(20)^2}{2 \times 10} = \frac{400}{20} = 20 \, \text{m}\]Consequently, the ball reaches a maximum height of \( 20 \, \text{m} \).