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

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 $H_{2\text{max}} = 16 \, \text{m}$.