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WAEC Physics 2015 Paper

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Question : 8

Total: 50

A body is projected with an initial velocity U at an angle O to the horizontal. The time taken by it to reach its maximum height is given by the expression

$\frac{2u \sin \theta}{g}$
$\frac{u^2 \sin^2 \theta}{2g}$
$\frac{u \sin \theta}{g}$
$\frac{u \sin \theta}{2g}$

Solution:

Concept:

At the maximum height, the vertical component of the projectile's velocity becomes zero.

Explanation:

The initial vertical component of velocity is

u \sin \theta

.

At maximum height, the final vertical velocity is

0

.

Using the first equation of motion for vertical motion:

v = u_y - gt

.

Substitute

v = 0

and

u_y = u \sin \theta

:

0 = u \sin \theta - gt

Rearranging gives:

gt = u \sin \theta

t = \frac{u \sin \theta}{g}

This is the time taken to reach the maximum height.

Answer:

C.

\frac{u \sin \theta}{g}

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