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 given by:

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

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

Step 1: Relationship between maximum heights
If the initial velocity is halved, i.e., \( u' = \frac{u}{2} \), the new maximum height \( H_{2\text{max}} \) can be expressed as:

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

Substitute \( u' = \frac{u}{2} \):

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

Step 2: Simplify the expression
Simplify \( \left( \frac{u}{2} \right)^2 \):

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

Thus:

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

Step 3: Calculate the new maximum height
Substitute \( H_{1\text{max}} = 64 \, \text{m} \):

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

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