A projectile is fired from the ground level with a velocity of 300ms-1 at an angle of a 30o to the horizontal.
Calculate the time taken to reach the maximum height?
[g = 10ms-2]
Calculate the time taken to reach the maximum height?
[g = 10ms-2]
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Correct Answer: Option D
Explanation:
The total time of flight is given as
\(T = \frac{2usin\theta}{g}\)
Hence, time to maximum height is half of the total time of flight,Â
t = \(\frac{u sin\theta}{g}\)
= \(\frac{300 \text {sin} 30}{10}\) = 15s
The total time of flight is given as
\(T = \frac{2usin\theta}{g}\)
Hence, time to maximum height is half of the total time of flight,Â
t = \(\frac{u sin\theta}{g}\)
= \(\frac{300 \text {sin} 30}{10}\) = 15s