There are two cliffs separated by 40 meters (horizontal displacement). An arrow is shot from a bow at an angle and a velocity of 90 m/s. The arrow takes 0.75 seconds to arrive at the other cliff. What is the angle at which the arrow was released? Neglect air resistance.
dx = vix*t + 0.5*ax*t2
Cos Θ = vix / v
The Attempt at a Solution
I used the equations and came up with 53.7°.
However, my question is why can I not solve for the initial velocity in the y direction using dy = viy + 0.5*ay*t2 (where 2 is half of the original time) and solve for the displacement the arrow traveled in the y, then use vfy2 = viy2 + 2ayd
Using those formulas and using SinΘ, I calculate 2.34°.
They both can't be correct. Why is the second one incorrect?
And let me add that I'm the teacher that created this question, so it is possible that my question sucks. As I'm grading my student's papers, I'm wondering about this alternate calculation and I can't seem to find the flaw in the work. But it's not what I was expecting.