Question

If the speed of projection of a projectile is twice its speed when it is at maximum height, then the angle of projection is

• A. 0°
• B. 30°
• C. 45°
• D. 60°

Hint:

Key concepts for projectile motion: \textbf{Initial velocity (u):} Has horizontal (u cos\( \theta \)) and vertical (u sin\( \theta \)) components. \textbf{Velocity at maximum height:} Only the horizontal component (u cos\( \theta \)) exists; the vertical component is zero. \textbf{Horizontal velocity:} Remains constant throughout the flight (ignoring air resistance). \textbf{Vertical velocity:} Changes due to gravity.

Answer 1

The Correct Option is D

Step 1: Define initial velocity and velocity at maximum height for a projectile.
Let the initial speed of projection be u, and the angle of projection with the horizontal be \( \theta \).
The initial velocity components are: Horizontal component (u\textsubscript{x}) = u cos\( \theta \) Vertical component (u\textsubscript{y}) = u sin\( \theta \) At the maximum height of a projectile's trajectory, the vertical component of velocity becomes zero (v\textsubscript{y} = 0). Only the horizontal component of velocity remains, which is constant throughout the projectile's flight (assuming no air resistance).
So, the speed at maximum height (v\textsubscript{max height}) = u\textsubscript{x} = u cos\( \theta \). Step 2: Use the given condition to set up an equation.
The problem states that the speed of projection (u) is twice its speed when it is at maximum height (u cos\( \theta \)).
So, \( \text{u} = 2 (\text{u cos}\theta) \) Step 3: Solve the equation for the angle of projection (\(\theta\)).
Divide both sides by u (assuming u \( \neq \) 0):
\( 1 = 2 \text{cos}\theta \) \( \text{cos}\theta = \frac{1}{2} \) To find \( \theta \), we take the inverse cosine: \( \theta = \text{cos\textsuperscript{-1}}(\frac{1}{2}) \) We know that cos(60°) = \( \frac{1}{2} \).
Therefore, \( \theta = 60 \)°.
Step 4: Select the correct option.
The angle of projection is 60°, which matches option (4).