If a projectile is shot at an angle of (5pi)/12 and at a velocity of 6 m/s, when will it reach its maximum height??

1 Answer
Feb 20, 2016

t_e~=0,59 s

Explanation:

"you can find the elapsed time to maximum height by: "
t_e=v_i*sin alpha /g
v_i:"initial velocity of object"
alpha:"angle of projectile :" alpha=(5pi)/12*180/pi=75^o
g:"acceleration of gravity"
t_e=(6*sin 75)/(9,81)
t_e~=0,59 s