Question

The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is ______ m.

Answer 1

Correct Answer: 16

The maximum height of a projectile is determined by the formula:

\[ H_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g}, \]

where \( u \) represents the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

Step 1: Analyzing the relationship between maximum heights
When the initial velocity is reduced by half, indicated as \( u' = \frac{u}{2} \), the subsequent maximum height, denoted as \( H_{2\text{max}} \), can be related to the initial maximum height \( H_{1\text{max}} \) as follows:

\[ \frac{H_{1\text{max}}}{H_{2\text{max}}} = \frac{u^2}{u'^2}. \]

Substituting the new initial velocity \( u' = \frac{u}{2} \) into the equation:

\[ \frac{H_{1\text{max}}}{H_{2\text{max}}} = \frac{u^2}{\left( \frac{u}{2} \right)^2}. \]

Step 2: Simplifying the derived expression
Simplifying the squared term \( \left( \frac{u}{2} \right)^2 \):

\[ \frac{H_{1\text{max}}}{H_{2\text{max}}} = \frac{u^2}{\frac{u^2}{4}} = 4. \]

This implies the following relationship for the new maximum height:

\[ H_{2\text{max}} = \frac{H_{1\text{max}}}{4}. \]

Step 3: Calculating the new maximum height
Using the given initial maximum height of \( H_{1\text{max}} = 64 \, \text{m} \):

\[ H_{2\text{max}} = \frac{64}{4} = 16 \, \text{m}. \]

Consequently, the projectile's new maximum height is \( H_{2\text{max}} = 16 \, \text{m} \).