Asked by Favour
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.
Answers
Answered by
oobleck
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
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
Answered by
Anonymous
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