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Here is the question:

A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no obstacles in its way. It is fired at an angle 33° with respect to the horizontal and with an initial speed of 36 m/s. Air resistance is negligible in this situation.

The same projectile was then fired in the same way toward a wall that is a horizontal distance 106.2 m from where the projectile was fired.

What was the height of the projectile when it hit the wall?

t = vx/dx = 106.2 m / 36cos(33degrees) = about 3.5 s

Peak is at 2 s: vfy = viy + at = 0 = 36sin(33degrees) - 9.8t

3.5 s - 2 s = 1.5 s

y = vyt + (1/2)(ay(t^2)) = 0 + (.5*9.81*1.51747^2) = 11.3 m

I got 11.3 m, and it is NOT the correct answer