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

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3-2.Motion in Plane

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

A

$5:4$

B

$5:2$

C

$5:1$

D

$10:1$

Solution

(a) $H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ and $T = \frac{{2u\sin \theta }}{g}$

So $\frac{H}{{{T^2}}} = \frac{{{u^2}{{\sin }^2}\theta /2g}}{{4{u^2}{{\sin }^2}\theta /{g^2}}} = \frac{g}{8} = \frac{5}{4}$

Standard 11

Physics

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