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 \( H \) reached by a projectile is determined using the vertical component of the initial velocity. Let's break down the solution:

1. Identify the vertical component of the initial velocity: The initial velocity can be divided into horizontal and vertical components. The vertical component \( u_y \) is given by: \[ u_y = u \sin \theta \]

2. Find the expression for maximum height: At maximum height, the vertical velocity becomes zero. Using the equation of motion: \[ v^2 = u^2 + 2as \] where \( v = 0 \), \( a = -g \), and \( s = H \) (the displacement or height), the equation becomes: \[ 0 = (u \sin \theta)^2 - 2gH \]

3. Solve for \( H \): Rearranging the equation to solve for \( H \), we get: \[ (u \sin \theta)^2 = 2gH \] \[ H = \frac{(u \sin \theta)^2}{2g} \]

Thus, the maximum height \( H \) reached by the projectile is \( \dfrac{u^2 \sin^2 \theta}{2g} \).

Answer 2

Step 1: Use the kinematic formula for maximum height. 
The vertical component of velocity is \( u_y = u \sin \theta \). At the maximum height, the vertical velocity becomes zero. 
Using the equation: \[ v^2 = u^2 - 2gH \quad \Rightarrow \quad 0 = (u \sin \theta)^2 - 2gH \] Step 2: Solve for \( H \). \[ H = \dfrac{u^2 \sin^2 \theta}{2g} \]