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

Concept: In projectile motion minimum velocity occurs at highest point. At highest point vertical component becomes zero. Hence minimum velocity equals horizontal component. \[ u\cos\theta=15 \] Maximum height formula: \[ H=\frac{u^2\sin^2\theta}{2g} \]

Step 1: Use height relation.
Given \[ 20=\frac{u^2\sin^2\theta}{20} \] \[ u^2\sin^2\theta=400 \] \[ u\sin\theta=20 \]

Step 2: Combine horizontal and vertical components.
We know \[ u\cos\theta=15 \] and \[ u\sin\theta=20 \]

Step 3: Find resultant velocity.
Using identity \[ u^2=(u\sin\theta)^2+(u\cos\theta)^2 \] \[ u^2=20^2+15^2 \] \[ u^2=625 \] \[ u=25\text{ ms}^{-1} \] Hence \[ \boxed{25\text{ ms}^{-1}} \]