Question

If minimum velocity of a projectile is \(15\text{ ms}^{-1}\) and maximum height reached is \(20\text{ m}\), then velocity of projection is (\(g=10\text{ ms}^{-2}\))

• A. \(35\text{ ms}^{-1}\)
• B. \(30\text{ ms}^{-1}\)
• C. \(20\text{ ms}^{-1}\)
• D. \(25\text{ ms}^{-1}\)

Hint:

For projectiles, minimum speed occurs at highest point and equals horizontal component.

Answer 1

The Correct Option is D

Step 1: Identify the minimum speed.
In projectile motion the speed is least at the top of the path, where only the horizontal component survives. So $u\cos\theta=15$ m/s.
Step 2: Use the maximum height.
The height formula is $H=\dfrac{u^2\sin^2\theta}{2g}=\dfrac{(u\sin\theta)^2}{2g}$.
Step 3: Solve for the vertical component.
$20=\dfrac{(u\sin\theta)^2}{2\cdot10}$, so $(u\sin\theta)^2=400$ and $u\sin\theta=20$ m/s.
Step 4: Collect the two components.
Horizontal: $u\cos\theta=15$. Vertical: $u\sin\theta=20$.
Step 5: Combine with Pythagoras.
$u^2=(u\sin\theta)^2+(u\cos\theta)^2=20^2+15^2=400+225=625$.
Step 6: Take the root.
$u=\sqrt{625}=25$ m/s.
\[ \boxed{25\ \text{ms}^{-1}} \]