if you give a golf ball an initial speed of V = 130 [miles per hour] = 58,1152 [meters per second] at an angle of α = 60 degrees,
a. how far will the ball travel assuming level ground?
Horizontal component of velocity Vh = V*cosα = 58,1152*cos60 = 58,1152/2 = 29,0576 m/s.
Vertical component of velocity Vv = V*sivα = 58,1152*sin60 = 58,1152*sqrt(3)/2 = 50,3292 m/s.
At the top point Vv-g*T = 0, where T is the time reached to get the peak height, so T = Vv/g = 50,3292/9.8 = 5,1356 s.
Peak height H = VvT - (gT^2)/2 = 50,3292*5,1356 - (9.8/2)*5,1356^2 = 129,2361 m.
Total time of flight is Tt = 2T = 2*5,1356 = 10,2712 s.
So, the distance is Tt*Vh = 10,2712*29,0576 = 298,4564 m.
b. how long will it take the ball to come back down to the ground?
As we already admitted, total time of flight Tt = 10,2712 s.
c. what is the peak height of the ball?
As we already admitted, peak height H = 129,2361 m.
d. how far will the ball travel if it is hit from a 20 m cliff?
We can submit that the ball has additional time to fall Ha = 20 m down. We can find this time Ta from equation
Vv*Ta+g*Ta^2 = Ha,
50,3292*Ta+9.8*Ta^2 = 20.
Solving this quadratic equation we obtain
Ta = 1,4825 s.
So, additional distance is
Da = Vv*Ta = 29,0576*1,4825 = 43,0779 m
and the total distance
Dt = Da+D = 43,0779+298,4564 = 341,5343 m.