Projection Under uniform Acceleration

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Question

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 $V_{o}$

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Solution

If angle of projection in one case is α, angle of projection in the other case is $(90^{o} - α)$
$\begin{array}{l} ∴ H_{1} = \frac{{V_{0}}^{2} \sin^{2} α}{2 g} and H_{2} = \frac{{V_{0}}^{2} \sin^{2} (90^{o} - α)}{2 g} = \frac{{V_{0}}^{2} \cos^{2} α}{2 g} \\ ∴ H_{1} + H_{2} = \frac{{V_{0}}^{2}}{2 g} \Rightarrow V_{0} = \sqrt{2 g (H_{1} + H_{2})} = \sqrt{2 \times 10 \times 90} m / s = 30 \sqrt{2} m / s \end{array}$

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

A ball is thrown at different angles with the same speed Vo 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=10m/s2, find Vo

If angle of projection in one case is α, angle of projection in the other case is(90o−α)∴H1=V02sin2α2g and H2=V02sin2(90o−α)2g=V02cos2α2g∴H1+ H2=V022g⇒V0=2g(H1+ H2)=2×10×90 m/s=302 m/s

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free study material

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 $V_{o}$