Question

If the minimum velocity of a projectile during its motion is \(40\,\text{m s}^{-1}\) and the ratio of its vertical and horizontal displacements at a time of \(2\,\text{s}\) is \(1:2\), then the angle of projection of the projectile is:

• A. \(\sin^{-1}(0.6)\)
• B. \(\cos^{-1}(0.6)\)
• C. \(\tan^{-1}(0.6)\)
• D. \(\sec^{-1}(0.6)\)

Hint:

In projectile motion, the minimum speed always equals the horizontal component: \[ v_{\min}=u\cos\theta \] because vertical velocity becomes zero at the highest point.

Answer 1

The Correct Option is A

Concept: In projectile motion, the horizontal component of velocity remains constant throughout the motion. The minimum speed occurs at the highest point where vertical velocity becomes zero. Thus: \[ v_{\min}=u\cos\theta \] Given: \[ u\cos\theta = 40 \] Also, \[ x = (u\cos\theta)t \] \[ y = (u\sin\theta)t - \frac12 gt^2 \]

Step 1: Use the displacement ratio condition. At \(t=2\,\text{s}\), \[ \frac{y}{x}=\frac{1}{2} \] Now, \[ x=(u\cos\theta)(2) \] \[ =40\times2 \] \[ =80\,\text{m} \] Vertical displacement: \[ y=(u\sin\theta)(2)-\frac12(10)(2)^2 \] \[ =2u\sin\theta-20 \] Using ratio: \[ \frac{2u\sin\theta-20}{80}=\frac12 \]

Step 2: Solve for \(u\sin\theta\). \[ 2u\sin\theta-20=40 \] \[ 2u\sin\theta=60 \] \[ u\sin\theta=30 \]

Step 3: Find angle of projection. We already know: \[ u\cos\theta=40 \] Hence, \[ \tan\theta=\frac{30}{40} \] \[ =\frac34 \] Using \(3-4-5\) triangle: \[ \sin\theta=\frac35=0.6 \] Therefore, \[ \boxed{\theta=\sin^{-1}(0.6)} \]