Question

A ball is projected with a velocity, 10 ms-1, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be:

• A. Zero
• B. 5√3ms^-1
• C. 5 ms^-1
• D. 10 ms^-1

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In projectile motion, the velocity of the projectile has two components: horizontal and vertical.
The horizontal component of velocity remains constant throughout the flight because there is no horizontal acceleration.
At the highest point, the vertical component of velocity becomes zero.
Therefore, the total speed at the highest point is simply the magnitude of the horizontal component of the initial velocity.
Key Formula or Approach:
If a body is projected with initial velocity \(u\) at an angle \(\theta\) with the horizontal:
\[ v_{\text{top}} = u_x = u \cos \theta \]
Step 2: Detailed Explanation:
1. Identify the angle with the horizontal:
Given: Initial velocity \(u = 10 \text{ ms}^{-1}\).
Angle with the vertical = \(60^\circ\).
Angle with the horizontal \(\theta = 90^\circ - 60^\circ = 30^\circ\).
2. Calculate horizontal component:
The speed at the highest point is \(u \cos \theta\).
\[ v_{\text{top}} = 10 \times \cos 30^\circ \]
Using the value \(\cos 30^\circ = \frac{\sqrt{3}}{2}\):
\[ v_{\text{top}} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ ms}^{-1} \]
Step 3: Final Answer:
The speed at the highest point is \(5\sqrt{3} \text{ ms}^{-1}\).