Question

A body is projected with a velocity of \(15\sqrt3\,\text{m s}^{-1}\) at an angle of \(60^\circ\) with the horizontal and another body is projected simultaneously from the same point in the same vertical plane with a velocity of \(40\,\text{m s}^{-1}\) at an angle of \(30^\circ\) with the horizontal. The time at which the velocity vectors of the two bodies will be in the same direction is:

• A. \(3.2\,\text{s}\)
• B. \(2.4\,\text{s}\)
• C. \(1.2\,\text{s}\)
• D. \(3.6\,\text{s}\)

Hint:

Whenever two velocity vectors are said to be parallel or in the same direction, equate their slopes: \[ \frac{v_y}{v_x} \] for both bodies.

Answer 1

The Correct Option is B

Concept: Two vectors are in the same direction if their slopes are equal. Velocity components for projectile motion are: \[ v_x=u\cos\theta \] \[ v_y=u\sin\theta-gt \]

Step 1: Find velocity components of first body. Given: \[ u_1=15\sqrt3,\qquad \theta_1=60^\circ \] Horizontal velocity: \[ v_{1x}=15\sqrt3 \times \frac12 \] \[ =\frac{15\sqrt3}{2} \] Vertical velocity: \[ v_{1y}=15\sqrt3\left(\frac{\sqrt3}{2}\right)-10t \] \[ =\frac{45}{2}-10t \]

Step 2: Find velocity components of second body. Given: \[ u_2=40,\qquad \theta_2=30^\circ \] Horizontal velocity: \[ v_{2x}=40\cos30^\circ \] \[ =20\sqrt3 \] Vertical velocity: \[ v_{2y}=40\sin30^\circ-10t \] \[ =20-10t \]

Step 3: Equate slopes of velocity vectors. \[ \frac{v_{1y}}{v_{1x}} = \frac{v_{2y}}{v_{2x}} \] \[ \frac{\frac{45}{2}-10t}{\frac{15\sqrt3}{2}} = \frac{20-10t}{20\sqrt3} \] Cross multiplying: \[ 20\left(\frac{45}{2}-10t\right) = 15(20-10t) \] \[ 450-200t = 300-150t \] \[ 150=50t \] \[ t=3 \] Checking carefully with simplification: \[ \frac{45-20t}{15\sqrt3} = \frac{20-10t}{20\sqrt3} \] \[ 20(45-20t)=15(20-10t) \] \[ 900-400t=300-150t \] \[ 600=250t \] \[ t=2.4\,\text{s} \] Hence, \[ \boxed{2.4\,\text{s}} \]