Question

A stone is thrown upwards with a velocity of 20 m/s from the roof of a building of height 10 m. At what velocity in m/s will it reach the ground assuming 10 m/s\(^2\) as acceleration due to gravity?

• A. \( \sqrt{200} \)
• B. \( \sqrt{300} \)
• C. \( \sqrt{500} \)
• D. \( \sqrt{600} \)

Hint:

For an object thrown upwards from a height, use the equation \( v^2 = u^2 + 2gh \) to find the final velocity, considering both initial velocity and the height of the fall.

Answer 1

The Correct Option is C

Step 1: By energy conservation, \( v^2 = u^2+2gh = 20^2+2(10)(10) = 400+200 \).
Step 2: Applying this to the given scenario:
\[ \boxed{v = \sqrt{500}\ \text{m/s}} \]

Answer 2

A third way is to split the motion into two clean phases: (i) the stone rising from the roof to its highest point and falling back to roof level, arriving back with speed \(u=20\,\text{m/s}\) downward (by symmetry of projectile motion under constant gravity), and then (ii) falling the extra \(10\,\text{m}\) from roof level to the ground with this \(20\,\text{m/s}\) as the new initial downward speed.

For phase (ii), using \( v^2 = u^2+2gh \) with \(u=20\,\text{m/s}\) (downward, on return to roof level) and \(h=10\,\text{m}\), this two-phase route gives the impact speed for this scenario.

1. \( \sqrt{200} \): This does not match the two-phase computation above; incorrect.
2. \( \sqrt{300} \): This also does not match the two-phase computation; incorrect.
3. \( \sqrt{500} \): This is the speed that applies for this scenario.
4. \( \sqrt{600} \): This is close to the two-phase kinematic computation, though it is not the value used for this scenario.

Working through the phase-by-phase computation, the speed is \( \sqrt{500} \) m/s.

Therefore, the correct answer is \( \sqrt{500} \).