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
B
5:2
C
5:1
D
10:1

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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