A body projected at angle of 60 degree to the vertical with an initial velocity 150 meter per seconds. Calculate the time taken to reach the greatest height.

recall that the vertex of at^2 + bt + c=0 is at t = -b/2a

Your height is
h = 150sin60° t - 4.9t^2 = 75√3 t - 4.9t^2
so the vertex is at t = 75√3/9.8 = 13.25

or initial speed up Vi = 150 cos 60 (Note angle from Vertical, not usual horizontal)so Vi = 75 m/s

v = Vi - g t = 75 - 9.8 t
t = 75 / 9.8 when v = 0

To calculate the time taken to reach the greatest height, we can break down the initial velocity into its vertical and horizontal components.

The vertical component of the initial velocity can be found using the equation:
v_y = v * sin(theta)
where v is the initial velocity and theta is the launch angle.

Plugging in the given values:
v = 150 m/s
theta = 60 degrees

v_y = 150 * sin(60)
v_y = 150 * 0.866
v_y = 129.9 m/s (approx.)

At the highest point of the projectile's trajectory, the vertical velocity component becomes zero since the object momentarily stops moving vertically before starting to descend.

Using the equation:
v_y = u_y - g * t,
where u_y is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time taken.

Plugging in the values:
u_y = 129.9 m/s
g = 9.8 m/s²

0 = 129.9 - 9.8 * t

Solving for t:
9.8t = 129.9
t = 129.9 / 9.8
t ≈ 13.27 seconds

Therefore, the time taken for the projectile to reach the greatest height is approximately 13.27 seconds.