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]
- 30s
- 20s
- 25s
-
15s
The correct answer is: 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,
t = \(\frac{u sin\theta}{g}\)
= \(\frac{300 \text {sin} 30}{10}\) = 15s
There is an explanation video available .