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 - Physics

MCQ

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

• 25 m

• 37 m

• 50 m

• 100 m

Solution

100 m
For the same range, $R \propto \sin\left( 2\theta \right)$ .

So, $\frac{R_1}{R_2} = \frac{\sin\left( 2 \theta_1 \right)}{\sin\left( 2 \theta_2 \right)}$

$\Rightarrow R_2 = 50 \times \frac{\sin\left( 90 \right)}{\sin\left( 30 \right)} = 100 \text{ m }$

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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 3 Rest and Motion: Kinematics
MCQ | Q 10 | Page 49