A projectile is shot at a velocity of 21 m/s and an angle of pi/8 . What is the projectile's peak height?

1 Answer
Mar 6, 2016

=62.1m

Explanation:

If the projectile is shot with velocity u at an angle of projection alpha with the horizontal, then it will have initial vertical component of velocity usinalpha and this velocity will becomes zero at maximum height H
So Equation of motion can be wrtten as
0^2=(u*sinalpha)^2-2*g*H
H=(u*sinalpha)^2/(2*g)=21^2sin^2(pi/8)/(2*9.8)=62.1m