Find Angle of Projectile: Golf Ball 50m High & Away

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To find the initial angle of a golf ball projected to a window 50m high and 50m away, the final vertical velocity is assumed to be zero at the maximum height. The vertical initial velocity is calculated to be 31.32 m/s, and the time of flight is determined to be 3.192 seconds. The horizontal initial velocity is found to be 15.66 m/s. Using these values, the initial angle is computed to be 63 degrees. The calculations and method are confirmed to be correct given the assumption that the ball hits the window at its maximum height.
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Homework Statement


A golf ball is hit from the ground into a building window 50m high and 50m away from starting point. Find the initial angle.

The Attempt at a Solution


I will work backwards for this question. I assume that the final y velocity (Vy) is 0m/s.

the first equation I do is: Vy^2= Viy^2 + 2ay(y)
I solve this equation for Viy and I get 31.32m/s.

Next I use the equation: Vy = Viy + ayt
I solve this equation for time and I get 3.192s.

Next I use d = vt and in respects to x. I solve for v (being the initial x velocity). For this I get Vix to be 15.66m/s.

Now I have initial x and y velocity and use inverse tan to solve for angle. I get the initial angle to be: 63 degrees.

Are my numbers and method correct?
 
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I believe my steps are correct.
 
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Why do you assume that the final vertical velocity is zero?
 
The final y velocity I assume to be zero. I forgot to add in a part where it said that it hit the window at its max height of motion.

Given these circumstances I am correct, no?
 
With the clarification, your solution is correct.
 
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