First slide

Projection Under uniform Acceleration

Question

The range of a projectile launched at an angle of $15^{o}$ with horizontal is 1.5 km. The range of projectile when launched at an angle of $45^{o}$ to the horizontal is

Moderate

A

1.5 km
B

3 km
C

6 km
D

0.7 km

Solution

$R_{15^{0}} = \frac{u^{2}}{g} \sin (2 \times 15^{0})$
$\Rightarrow 1.5 = \frac{u^{2}}{g} \sin 30^{0}$
$\Rightarrow \frac{u^{2}}{g} = \frac{1.5}{\sin 30^{0}} = 1.5 \times 2 = 3 km$
and $R_{45^{0}} = \frac{u^{2}}{g} \sin (2 \times 45^{0}) = \frac{u^{2}}{g} = 3 km$

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