Question

A ball is thrown upwards with a velocity of \(20\ \text{m/s}\); find the maximum height reached \((g = 10\ \text{m/s}^2)\).

• A. \(10\) metres
• B. \(20\) metres
• C. \(30\) metres
• D. \(40\) metres

Hint:

For maximum height problems, remember that the velocity becomes zero at the top. So directly use \( v^2 = u^2 - 2gh \).

Answer 1

The Correct Option is B

Concept: For motion under constant acceleration, we use the kinematic equation \[ v^2 = u^2 - 2gh \] where \(u\) = initial velocity, \(v\) = final velocity, \(g\) = acceleration due to gravity, \(h\) = maximum height. At the highest point, the velocity becomes zero.

Step 1: Substitute the given values. Initial velocity: \[ u = 20\ \text{m/s} \] At maximum height: \[ v = 0 \] Using \[ v^2 = u^2 - 2gh \] \[ 0 = (20)^2 - 2(10)h \]

Step 2: Solve for \(h\). \[ 0 = 400 - 20h \] \[ 20h = 400 \] \[ h = 20\ \text{m} \] Thus, the maximum height reached is \(20\) metres.