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Ok, so since we know the angle of "takeoff" and the total distance and time we can find the velocity, now the issue with this velocity is that it is the vertical velocity, not the velocity in the y-direction. However, if you've taken geometry you probably know about SOH-CAH-TOA, and you were probably taught to do this problem using triangles. Since you have the angle, and the hypotenuse (the vertical velocity you solved for) then you can find the x and y parts that make up the vertical velocity. Now I'm a bit confused as to wether you are solving for the distance it goes up or the total distance it goes overall.
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Answer:Explanation:Maximum height = U²sin²angle/2gWhere u is velocityAngle is in degreeg is acceleration due to gravityTime of flight = 2usinangle/gRange = U²sine2 x angle/g
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