Question

A projectile is fired with an initial velocity of 20 m/s at an angle of 30\(^\circ\) with the horizontal. Calculate the maximum height reached by the projectile.

• A. 10 m
• B. 5.1 m
• C. 20 m
• D. 25 m

Hint:

The maximum height of a projectile depends on the square of the initial velocity and the sine of the angle. The greater the initial velocity and angle, the higher the projectile will go.

Answer 1

The Correct Option is B

Given:

• Initial velocity: \( v_0 = 20 \, \text{m/s} \)
• Angle of projection: \( \theta = 30^\circ \)
• Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \)

Step 1: Calculate the vertical component of the initial velocity

The vertical component of the initial velocity is given by: \[ v_{0y} = v_0 \sin \theta \] Substituting the given values: \[ v_{0y} = 20 \, \text{m/s} \times \sin(30^\circ) = 20 \, \text{m/s} \times 0.5 = 10 \, \text{m/s} \]

Step 2: Use the formula for maximum height

The formula for the maximum height reached by a projectile is: \[ H = \frac{v_{0y}^2}{2g} \] Substituting the values: \[ H = \frac{(10 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} = \frac{100}{19.6} = 5.1 \, \text{m} \]

Step 3: Final Answer

The maximum height reached by the projectile is \( \boxed{5.1 \, \text{m}} \).