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Question

For a projectile, the ratio of maximum height reached to the square of the time of flight is $$(g=10  ms^{-2})$$


A
5:4
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B
5:2
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C
5:1
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D
10:1
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Solution

The correct option is A $$5:4$$
$$H_{max} = \dfrac{u^{2}\sin^{2} \theta}{2g}$$
$$T = \dfrac{2u \sin \theta}{g}$$
$$\displaystyle\dfrac{H_{max}}{T^{2}}=  \dfrac{(u^{2}\sin^{2} \theta)}{2g} \frac{g^{2}}{4u^{2}\sin^{2} \theta}= \dfrac{g}{8} = \dfrac{10}{8} = \dfrac{5}{4}$$

Physics

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