An object is protected with a velocity of 100m -1 at an angle of 60
PHYSICS
WAEC 1998
An object is protected with a velocity of 100m-1 at an angle of 60o 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
Correct Answer: B. 8.7s
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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