If a projectile is shot at an angle of pi/6 and at a velocity of 18 m/s, when will it reach its maximum height??

2 Answers
Mar 30, 2016

Time of reaching at maximum height
t =(usinalpha)/g=(18*sin(pi/6))/9.8=0.91s

Mar 30, 2016

t~=0,92 " s"

Explanation:

g=9,82 " m/s^2

v_i=18" "m/s

alpha=pi/6

sin alpha=0,5

t=(v_i*sin alpha)/g

t=(18*0,5)/(9,81)

t=9/(9,81)

t~=0,92 " s"