A ball is thrown at different angles with the same speed $V_{o}$ from the same point on ground and it has same range in both cases. If the sum of maximum heights attained by the ball is 90m and $g = 10 m / s^{2}$ , find $V_{o}$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

60m/s

$30 \sqrt{2} m / s$

$20 \sqrt{2} m / s$

$40 \sqrt{2} m / s$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

If angle of projection in one case is α, angle of projection in the other case is $(90^{o} - α)$

$\begin{array}{l} ∴ H_{1} = \frac{{V_{0}}^{2} \sin^{2} α}{2 g} and H_{2} = \frac{{V_{0}}^{2} \sin^{2} (90^{o} - α)}{2 g} = \frac{{V_{0}}^{2} \cos^{2} α}{2 g} \\ ∴ H_{1} + H_{2} = \frac{{V_{0}}^{2}}{2 g} \Rightarrow V_{0} = \sqrt{2 g (H_{1} + H_{2})} = \sqrt{2 \times 10 \times 90} m / s = 30 \sqrt{2} m / s \end{array}$