Question

Two projectiles \(P\) and \(Q\) thrown with velocities \(v\) and \(\dfrac{v}{2}\) respectively have the same range. If \(Q\) is thrown at an angle of \(15^\circ\) to the horizontal, \(P\) must be thrown at an angle of

• A. \(30^\circ\)
• B. \(\dfrac{1}{2}\sin^{-1}\left(\dfrac{1}{8}\right)\)
• C. \(\dfrac{1}{4}\sin^{-1}\left(\dfrac{1}{8}\right)\)
• D. \(60^\circ\)
• E. \(45^\circ\)

Hint:

If two projectiles have the same range, equate: \[ u_1^2\sin2\theta_1=u_2^2\sin2\theta_2 \]

Answer 1

The Correct Option is B

Range: \[ R=\frac{u^2\sin2\theta}{g} \] Same range means: \[ \frac{v^2\sin2\theta_P}{g} = \frac{(v/2)^2\sin30^\circ}{g} \] \[ v^2\sin2\theta_P = \frac{v^2}{4}\cdot \frac{1}{2} \] \[ \sin2\theta_P=\frac{1}{8} \] So, \[ 2\theta_P=\sin^{-1}\left(\frac{1}{8}\right) \] \[ \theta_P=\frac{1}{2}\sin^{-1}\left(\frac{1}{8}\right) \] Hence, \[ \boxed{(B)\ \frac{1}{2}\sin^{-1}\left(\frac{1}{8}\right)} \]