Question

A stone is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of \( 10 \, \text{m/s}^2 \), with what velocity and after what time will it strike the ground?

• A. 20 m/s, 2 s
• B. 10 m/s, 20 s
• C. 10 m/s, 2 s
• D. 20 m/s, 20 s

Hint:

Use the equations of motion to solve problems involving uniform acceleration. \( v^2 = u^2 + 2 a h \) helps find the final velocity, and \( v = u + at \) helps find the time.

Answer 1

The Correct Option is A

We are given: - Initial velocity \( u = 0 \, \text{m/s} \) (since the stone is gently dropped) - Acceleration \( a = 10 \, \text{m/s}^2 \) - Height \( h = 20 \, \text{m} \) Using the second equation of motion: \[ v^2 = u^2 + 2 a h \] Substituting the known values: \[ v^2 = 0 + 2 \times 10 \times 20 = 400 \] \[ v = \sqrt{400} = 20 \, \text{m/s} \] Now, using the first equation of motion to find the time: \[ v = u + at \] \[ 20 = 0 + 10 \times t \] \[ t = \frac{20}{10} = 2 \, \text{s} \] Thus, the velocity is \( 20 \, \text{m/s} \) and the time taken is 2 seconds.