Question

A particle is thrown with a speed \(v\) from a point \(O\) at an angle \(\theta\) with the horizontal plane such that it passes through the point \(P\) at a height of \(1\,\text{m}\) and horizontal distance of \(5\,\text{m}\) from \(O\), as shown in the figure. If acceleration due to gravity is \(g\,\text{m s}^{-2}\), then the correct statement(s) is/are:

• A. If \(\theta=45^\circ\), then \[ v=\frac{5\sqrt g}{2}\,\mathrm{m\,s^{-1}} \]
• B. If \(\theta=45^\circ\), the particle reaches its maximum height before it reaches \(P\)
• C. If \(\theta=30^\circ\), the particle reaches its maximum height after reaching \(P\)
• D. If \(\theta=\tan^{-1}\left(\frac15\right)\), then \[ v=125\sqrt g\,\mathrm{m\,s^{-1}} \]

Hint:

Projectile trajectory: \[ y=x\tan\theta-\frac{gx^2}{2v^2\cos^2\theta} \] Horizontal coordinate of maximum height: \[ x_H=\frac{v^2\sin2\theta}{2g} \]

Answer 1

The Correct Option is A

Step 1: Use projectile equation.
Trajectory equation: \[ y=x\tan\theta-\frac{gx^2}{2v^2\cos^2\theta} \] Given: \[ x=5,\qquad y=1 \] Thus: \[ 1=5\tan\theta-\frac{25g}{2v^2\cos^2\theta} \]

Step 2: Check option (A).
For: \[ \theta=45^\circ \] \[ \tan45^\circ=1, \qquad \cos^245^\circ=\frac12 \] Substitute: \[ 1=5-\frac{25g}{v^2} \] \[ \frac{25g}{v^2}=4 \] \[ v^2=\frac{25g}{4} \] \[ v=\frac{5\sqrt g}{2} \] Hence: \[ \boxed{\mathrm{(A)\ is\ correct}} \]

Step 3: Check option (B).
Horizontal distance of highest point: \[ x_H=\frac{v^2\sin2\theta}{2g} \] For: \[ \theta=45^\circ \] \[ \sin2\theta=1 \] Using: \[ v^2=\frac{25g}{4} \] \[ x_H=\frac{25g}{8g} =\frac{25}{8} \] \[ =3.125\,\text{m} \] Since: \[ 3.125<5 \] particle reaches maximum height before point \(P\). Hence: \[ \boxed{\mathrm{(B)\ is\ correct}} \]

Step 4: Check option (C).
For: \[ \theta=30^\circ \] Projectile equation: \[ 1=5\left(\frac1{\sqrt3}\right)-\frac{25g}{2v^2\cdot\frac34} \] \[ 1=\frac5{\sqrt3}-\frac{50g}{3v^2} \] \[ \frac{50g}{3v^2}=\frac5{\sqrt3}-1 \] Now: \[ x_H=\frac{v^2\sin60^\circ}{2g} \] \[ =\frac{v^2\sqrt3}{4g} \] Substitute: \[ v^2=\frac{50g}{3\left(\frac5{\sqrt3}-1\right)} \] This gives: \[ x_H>5 \] Thus maximum height is reached after crossing \(P\). Hence: \[ \boxed{\mathrm{(C)\ is\ correct}} \]

Step 5: Check option (D).
For: \[ \tan\theta=\frac15 \] \[ \cos^2\theta=\frac{25}{26} \] Substitute in projectile equation: \[ 1=5\left(\frac15\right)-\frac{25g}{2v^2\cdot\frac{25}{26}} \] \[ 1=1-\frac{13g}{v^2} \] Thus: \[ \frac{13g}{v^2}=0 \] Impossible. Hence: \[ \boxed{\mathrm{(D)\ is\ incorrect}} \]

Step 6: Identify the correct statements.
Therefore: \[ \boxed{\mathrm{(A)\ and\ (B)\ and\ (C)}} \]