If a projectile has a launching angle of 52 degrees about the horizontal and an initial speed of 18 m/s, what is the highest barrier that the projectile can clear?

Answer 1

Vo = 18m/s @ 52 deg.

Vo(hor) = 18cos52 = 11.08m/s.

Vo(ver) = 18sin52 = 14.18m/2,up.

Vf^2 = Vo^2 + 2gd = 0 @ max height.
(14.18)^2 + 2(-9.8)d = 0,
201.07 -19.6d = 0,
d = 10.3m. = max barrier that can be
cleared.