A football is kicked at an angle of 45° to the horizontal over a defense line up with a velocity of 15\(ms^{-1}\). Calculate the magnitudeof horizontal velocity of the ball at its highest point [Neglect friction, g = 10\(ms^{-2}\)].
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Correct Answer: Option C
Explanation:
The horizontal component of velocity, \(V_{x}\), is unaffected by the acceleration of free fall and remains constant.
\(V_{x} = U_{x} = Ucos\theta\)
= \(15 \times cos45 = 15\times 0.7071\)
\(\approxeq 10.6ms^{-1}\)
The horizontal component of velocity, \(V_{x}\), is unaffected by the acceleration of free fall and remains constant.
\(V_{x} = U_{x} = Ucos\theta\)
= \(15 \times cos45 = 15\times 0.7071\)
\(\approxeq 10.6ms^{-1}\)