Question

A 0.2 kg ball is thrown vertically upwards with an initial velocity of \( 10 \, \text{m/s} \). What is the maximum height reached by the ball? (Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \))

• A. \( 5 \, \text{m} \)
• B. \( 10 \, \text{m} \)
• C. \( 20 \, \text{m} \)
• D. \( 2 \, \text{m} \)

Hint:

When an object is thrown upwards, use the kinematic equation \( v^2 = u^2 - 2gh \) to find the maximum height. At the maximum height, the final velocity is zero.

Answer 1

The Correct Option is A

Step 1: Determine maximum height using kinematic equation At its peak, the projectile's velocity is zero (\( v = 0 \, \text{m/s} \)). The relevant equation for vertical motion is:\[v^2 = u^2 - 2 g h\]Where:- \( v \) denotes final velocity.- \( u \) represents initial velocity.- \( g \) is the acceleration due to gravity.- \( h \) signifies maximum height.Given values:- \( u = 10 \, \text{m/s} \)- \( v = 0 \, \text{m/s} \)- \( g = 9.8 \, \text{m/s}^2 \)Re-arranging the equation to solve for \( h \):\[0 = (10)^2 - 2 \times 9.8 \times h\]\[h = \frac{(10)^2}{2 \times 9.8} = \frac{100}{19.6} \approx 5 \, \text{m}\]Answer: The projectile reaches a maximum height of approximately \( 5 \, \text{m} \). This corresponds to option (1).