A golfer stands 480ft (160yd) horizontally from the hole and 60 ft above the hol
ID: 2880515 • Letter: A
Question
A golfer stands 480ft (160yd) horizontally from the hole and 60
ft above the hole (see figure.) Assuming the ball is hit with an initial velocity of 120
ft/s, at what angle (or angles) should it be hit to land in the hole? Assume the path of the ball lies in a plane.
A golfer stands 480 ft (160 yd) horizontally from the hole and 60 ft above the hole (see figure.) Assuming the ball is hit with an initial 60 ft velocity of 120 ft/s, at what angle (or angles) t should it be hit to land in the hole? Assume the path of the ball lies in a plane. 480 ft (160 yd) The ball should be hit at an angle (or angles) of degrees (Type an integer or decimal rounded to two decimal places as needed. Use a comma to separate answers as needed.)Explanation / Answer
x_o= 480 ft
y_o=60 ft
v_o= 120 ft/s
g= 32 ft/s^2
x-xo=vo(cosx)t
480=120(cosx)t;
t=4/(cosx)
y-yo=(vo*sinx)t-1/2*g*t²
-60=120*sinx*4/(cosx)-16*16/(cos²x)
0=480sinx*cosx-256+60cos²x
x= 6.9°