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WAEC Physics 2018 Paper

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Question : 14

Total: 50

A football is kicked at an angle of 45° to the horizontal over a defense line up with a velocity of 15

\mathrm{ms}^{-1}

. Calculate the magnitudeof horizontal velocity of the ball at its highest point [Neglect friction, g = 10

\mathrm{ms}^{-2}

].

0.0 $\mathrm{ms}^{-1}$
5.6 $\mathrm{ms}^{-1}$
10.6 $\mathrm{ms}^{-1}$
15.0 $\mathrm{ms}^{-1}$

Solution:

The horizontal component of velocity, $V_{x}$ , is unaffected by the acceleration of free fall and remains constant.

$V_{x} = U_{x} = U\cos\theta$

= $15 \times \cos 45 = 15 \times 0.7071$

$\approx\text{eq} 10.6\mathrm{ms}^{-1}$

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