OnlineCBT
Physics · MOTION IN A STRAIGHT LINE
हिन्दीA ball is thrown vertically upwards with a velocity of \( 20.0\text{ m s}^{-1} \) from the top of a building of height \( 25.0\text{ m} \) above the ground. Taking \( g = 10.0\text{ m s}^{-2} \), what is the maximum height reached by the ball above the ground level?
Correct answer · Option 3
\( 45.0\text{ m} \)
Explanation
The height risen above launch point is \( h = \frac{v_0^2}{2g} = \frac{20.0^2}{2 \times 10.0} = 20.0\text{ m} \). Total height above ground = \( 25.0\text{ m} + 20.0\text{ m} = 45.0\text{ m} \).
