First slide

Projection Under uniform Acceleration

Question

A body is projected at $60^{o}$ with ground. It covers a horizontal distance of 100 m. If the same body is projected at $60^{0}$ with vertical with same velocity, the new range is

Moderate

A

50 m
B

100 m
C

200 m
D

150 m

Solution

Range, R = $\frac{u^{2} \sin 2 θ}{g}$
For same v, $R \propto 2 θ$
For $θ = 60^{0}$ with horizontal, R is given as 100 m when R $\propto \sin 120^{0}$ --------(i)
For $θ = 60^{0} with vertical$
$R^{'} \propto \sin 2 (90^{0} - 60^{0})$
i.e., $R^{'} \propto \sin 60^{0}$ ---------------(ii)
Using (i) and (ii)
$\frac{R^{'}}{R} = \frac{\sin 60^{0}}{\sin 120^{0}} = \frac{1}{1}$
i.e., $R^{'} = R = 100 m$

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