For a projectile ratio of maximum height reached to square of flight time is

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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

$5 : 4$

$5 : 2$

$5 : 1$

$10 : 1$

Solution

$\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}$

and $\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}$

$\frac{\mathrm{H}}{\mathrm{T}^{2}}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}} \times \frac{\mathrm{g}^{2}}{4 \mathrm{u}^{2} \sin \theta}=\frac{\mathrm{g}}{8}=\frac{10}{8}=\frac{5}{4}$

Standard 11

Physics

In projectile motion, the modulus of rate of change of velocity

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A particle $A$ is projected vertically upwards. Another identical particle $B$ is projected at an angle of $45^o $ . Both reach the same height. The ratio of the initial kinetic energy of $A$ to that of $B$ is

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(AIIMS-2001)

A cricketer can throw a ball to a maximum horizontal distance of $100\, m .$ The speed with which he throws the ball is (to the nearest integer) (in $ms ^{-1}$)

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(AIIMS-2019)

For a projectile the ratio of maximum height reached to the square of flight time is

Solution

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