Question

A projectile is fired with an initial velocity \( u \) at an angle \( \theta \) to the horizontal. The time of flight is \( T \). What is the maximum height \( H \) reached by the projectile?

• A. \( \dfrac{u^2 \sin^2 \theta}{2g} \)
• B. \( \dfrac{u^2 \sin 2\theta}{2g} \)
• C. \( \dfrac{u^2 \sin^2 \theta}{g} \)
• D. \( \dfrac{u^2 \sin \theta}{2g} \)

Hint:

Maximum height in projectile motion is determined by vertical velocity component: \[ H = \dfrac{(u \sin \theta)^2}{2g} \] Use this when the object reaches the topmost point (vertical velocity becomes zero).

Answer 1

The Correct Option is A

The maximum height \( H \) attained by a projectile is calculated using the vertical component of its initial velocity. The process is as follows:

1. Determine the vertical component of initial velocity: The initial velocity can be decomposed into horizontal and vertical elements. The vertical component, \( u_y \), is derived from: \[ u_y = u \sin \theta \]

2. Derive the formula for maximum height: At its apex, the projectile's vertical velocity is momentarily zero. Applying the equation of motion \( v^2 = u^2 + 2as \), with \( v = 0 \) (final vertical velocity), \( a = -g \) (acceleration due to gravity), and \( s = H \) (vertical displacement), yields: \[ 0 = (u \sin \theta)^2 - 2gH \]

3. Isolate \( H \): Rearranging the equation to solve for the maximum height \( H \): \[ (u \sin \theta)^2 = 2gH \] \[ H = \frac{(u \sin \theta)^2}{2g} \]

Consequently, the maximum height \( H \) of the projectile is given by the expression \( \dfrac{u^2 \sin^2 \theta}{2g} \).