Explanation:
It is given that,
The angle of projection is 60 degrees
Initial velocity of the ball is 120 m/s
We need to find the time taken to get to the maximum height and the time of flight.
Time taken to reach the maximum height is given by :
[tex]T=\dfrac{u^2\sin^2\theta}{2g}[/tex]
g is acceleration due to gravity
[tex]T=\dfrac{(120)^2\times \sin^2(60)}{2\times 10}\\\\T=540\ s[/tex]
(ii) Time of flight,
[tex]t=\dfrac{2u\sin\theta}{g}[/tex]
So,
[tex]t=\dfrac{2\times 120\times \sin(60)}{10}\\\\t=20.78\ s[/tex]
Hence, this is the required solution.
