A projectile is shot at a velocity of 3 m/s and an angle of pi/8 . What is the projectile's peak height?

1 Answer
Feb 3, 2016

h_(peak)=0,00888 "meters"

Explanation:

"the formula needed to solve this problem is:"
h_(peak)=(v_i^2*sin^2 theta/(2*g))
v_i=3 m/s
theta =180/cancel(pi)*cancel(pi)/8
theta =180/8
sin theta=0,13917310096
sin^2 theta=0,0193691520308
h_(peak)=3^2*(0,0193691520308)/(2*9,81)
h_(peak)=9*(0,0193691520308)/(19,62)
h_(peak)=0,00888 "meters"