Question

The horizontal range of a projectile fired at an angle of 15⁰ is 50 m. If it is fired with the same speed at
an angle of 45°, its range will be

Moderate

Solution

Given, $θ = 15^{o}$ and R = 50m
Range, $R = \frac{u^{2} \sin 2 θ}{g}$
Putting all the given values in the formula, we get
$R = 50 m = \frac{u^{2} \sin (2 \times 15 °)}{9}$
$\Rightarrow 50 \times g = u^{2} \sin 30 ° = u^{2} \times \frac{1}{2} \Rightarrow 50 \times g \times 2 = u^{2}$
$\Rightarrow u^{2} = 50 \times 9.8 \times 2 = 980$
$\Rightarrow u = \sqrt{980}$ m/s =14 $\sqrt{5} m / s$

$For θ = 45 °, R = \frac{u^{2} \sin 2 \times 45 °}{g} = \frac{u^{2}}{g}$ $(∵ \sin 90 ° = 1)$
$\Rightarrow R' = \frac{(14 \sqrt{5})^{2}}{9} = \frac{14 \times 14 \times 5}{9.8} = 100 m$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME