Test Index

WAEC Physics 2023 Paper

© africaexams.com

Question : 26

Total: 50

The ratio of maximum range to maximum height of a projectile is 4 . Determine the angle of projection.

$30^{\circ}$
$60^{\circ}$
$45^{0}$
$0^{0}$

Solution:

Concept:

For a projectile, the maximum range and maximum height are related through the angle of projection.

Explanation:

Maximum range,

R = \frac{u^{2}\sin 2\theta}{g}

Maximum height,

H = \frac{u^{2}\sin^{2}\theta}{2g}

Given

\frac{R}{H} = 4

Substitute the expressions:

\frac{\frac{u^{2}\sin 2\theta}{g}}{\frac{u^{2}\sin^{2}\theta}{2g}} = 4

\frac{2\sin 2\theta}{\sin^{2}\theta} = 4

Since

\sin 2\theta = 2\sin\theta \cos\theta

, we get:

\frac{4\sin\theta \cos\theta}{\sin^{2}\theta} = 4

\frac{\cos\theta}{\sin\theta} = 1

\cot\theta = 1

\theta = 45^{\circ}

Answer:

The angle of projection is

45^{\circ}

, which is option C.

© africaexams.com

Go to Question:

Prev Question

Next Question