An object is protected with a velocity of 100m-1 at an angle of 60o to...

WAEC 1998

An object is protected with a velocity of 100m^-1 at an angle of 60^o to the vertical. Calculate the time taken by the object to reach the highest point (Take g as 10ms^-2

A. 5.0s
B. 8.7s
C. 10.0s
D. 17.3s
E. 20.0s

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Correct Answer: Option B

Explanation

t = \(\frac{T}{2} = \frac{1}{2} = \frac{2u \sin \theta}{g} = \frac{100 \times \sin 60}{10}\)

10 sin60 = 10 x 0.8660

= 8.7

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An object is protected with a velocity of 100m-1 at an angle of 60o to...

Correct Answer: Option B

Explanation

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