The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be
(a) 25 m
(b) 37 m
(c) 50 m
(d) 100 m

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Solution

(d) 100 m

For the same u range, $R \propto \sin (2 θ)$ .
So,
$\frac{R_{1}}{R_{2}} = \frac{\sin (2 θ_{1})}{\sin (2 θ_{2})}$
$\Rightarrow R_{2} = 50 \times \frac{\sin (90)}{\sin (30)} = 100 m$

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The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be (a) 25 m (b) 37 m (c) 50 m (d) 100 m

(d) 100 m For the same u range, R∝sin2θ. So, R1R2=sin2θ1sin2θ2 ⇒R2=50×sin90sin30=100 m

The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be
(a) 25 m
(b) 37 m
(c) 50 m
(d) 100 m

(d) 100 m

For the same u range, $R \propto \sin (2 θ)$ .
So,
$\frac{R_{1}}{R_{2}} = \frac{\sin (2 θ_{1})}{\sin (2 θ_{2})}$
$\Rightarrow R_{2} = 50 \times \frac{\sin (90)}{\sin (30)} = 100 m$