A projectile is shot at angle 60 degrees above the horizonta

A projectile is shot at angle 60° above the horizontal with initial velocity 200 ft /sec. It passes through a hole in a vertical fence at a height y above the ground. As it passes through the hole it is traveling at 40° below the horizontal. 672 / (a) Find the time from O to H. 60° (b) Find height y

Solution

in horizontal, there is no acceleraton.
hence horizontal velocity will remains constant.

vx = 200cos60 = 100 ft/s

in vertical,

vy = vi + at = (200 sin60) + (-32.174 t) = 173.205 - 32.174t (this will be negative velocity)

|vy| = 32.174t - 173.205

and tan40 = vy / vx

0.839 = (32.174t - 173.205) / 100

t = 8 sec

y = uy*t + a*t^2 /2

y = (100 sin60 * 8) + (-32.174 * 8^2 /2 )

y = 336.75 ft

The variation is because the value of g used.

I used g = 32.174 ft/s^2.

plug in value given in your text then answer will be exact.