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

Step 1: Use the minimum speed clue.
In projectile motion the horizontal speed never changes. The vertical speed becomes zero at the top. So the smallest speed of the whole flight is just the horizontal part. This means \[ u\cos\theta = v_{\min} = 40 \]

Step 2: Write horizontal displacement at 2 seconds.
Horizontal motion has no acceleration, so distance is speed times time. \[ x = (u\cos\theta)\,t = 40 \times 2 = 80\,\text{m} \]
Step 3: Write vertical displacement at 2 seconds.
Vertical motion is slowed by gravity. \[ y = (u\sin\theta)t - \tfrac12 g t^2 = 2u\sin\theta - 20 \]
Step 4: Apply the given ratio.
We are told vertical to horizontal is $1:2$, so $y = \tfrac12 x = 40$. \[ 2u\sin\theta - 20 = 40 \] Solving gives \[ u\sin\theta = 30 \]
Step 5: Find the angle from the two parts.
We now have both pieces of the launch velocity. Divide them. \[ \tan\theta = \frac{u\sin\theta}{u\cos\theta} = \frac{30}{40} = \frac34 \]
Step 6: Read off sine of the angle.
A right triangle with sides $3$, $4$, $5$ matches $\tan\theta = \tfrac34$. So the opposite over hypotenuse is \[ \sin\theta = \frac35 = 0.6 \] \[ \boxed{\theta = \sin^{-1}(0.6)} \]