First slide

Projection Under uniform Acceleration

Question

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

Moderate

A

5:4
B

5:2
C

5:1
D

10 :1

Solution

$\large \frac{H}{{{T^2}}} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} \times \frac{{{g^2}}}{{4{u^2}{{\sin }^2}\theta }}\left( {T = \frac{{2u\sin \theta }}{g}} \right)$
$\large = \frac{g}{8} = \frac{{10}}{8} = \frac{5}{4}$

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