Question

A projectile is fired with an initial speed of 40 m/s at an angle of 30° with the horizontal. What is the maximum height it reaches?

• A. 10 m
• B. 15 m
• C. 20 m
• D. 40 m

Hint:

Maximum height depends on vertical component: \(H = \frac{(u \sin \theta)^2}{2g}\).

Answer 1

The Correct Option is C

Given: Initial speed, \( u = 40\, \text{m/s} \), Angle, \( \theta = 30^\circ \), Gravity, \( g = 9.8\, \text{m/s}^2 \). Vertical component of velocity: \[ u_y = u \sin \theta = 40 \times 0.5 = 20\, \text{m/s} \] Maximum height: \[ H = \frac{u_y^2}{2g} = \frac{20^2}{2 \times 9.8} = \frac{400}{19.6} = 20.41\, \text{m} \] Nearest option is 20 m.