First slide

Projection Under uniform Acceleration

Question

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

Easy

A

5 : 4
B

5 : 2
C

5 : 1
D

10 : 1

Solution

$H = \frac{U^{2} \sin^{2} θ}{2 g} T^{2} = \frac{4 U^{2} \sin^{2} θ}{g^{2}} \frac{H}{T^{2}} = \frac{g}{8} = \frac{5}{4} .$

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