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
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.
2 answers
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
v = Vi - g t = 75 - 9.8 t
t = 75 / 9.8 when v = 0