Question:medium

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.

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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.
Updated On: Nov 26, 2025
  • 10 m
  • 5.1 m 
     

  • 20 m
  • 25 m
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The Correct Option is B

Solution and Explanation

Given:

  • Initial velocity: \( v_0 = 20 \, \text{m/s} \)
  • Projection angle: \( \theta = 30^\circ \)
  • Gravitational acceleration: \( g = 9.8 \, \text{m/s}^2 \)

Step 1: Calculate initial vertical velocity

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

Step 2: Determine maximum height

The maximum height \( H \) is calculated using: \[ H = \frac{v_{0y}^2}{2g} \] 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: Result

The projectile achieves a maximum height of \( \boxed{5.1 \, \text{m}} \).

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