Question

After a golf ball is hit it takes off with an initial speed of 28.7 m/s and at an angle of 39.5° with respect to the horizontal. The golf field is flat and horizontal. Neglecting air resistance how far will the golf ball fly?

How high will the golf ball rise?

How much time will the ball spend in the air?

How far would the ball fly if the initial speed was doubled?

How much time would the ball spend in the air in this second case?

Answer 1

Vh = 28.7 * cos(39.5) = 22.15 m/s
Vy = 28.7 * sin(39.5) =  18.25 m/s

At max height , Vertical velocity = 0
Vf^2 = Vy^2 + 2*a*h
0 = 18.25^2 - 2*9.8*h
h = 17.0 m

Time taken to reach max height,
Vf = Vy + a*t
t = 18.25/9.8
t = 1.86 s

Total time will the ball spend in the air, = 2*t = 2 * 1.86 s
t = 3.72 s

If initial speed is doubled,
Vh = 2 * 22.15 = 44.3 m/s
Vy = 2 * 18.25 = 36.5 m/s

time would the ball spend in the air in this second case,
t = 2 * Vy/g
t = 2 * 36.5/9.8
t = 7.45 s

How far would the ball fly if the initial speed was doubled?
Sx = 44.3 * 7.45 m
Sx = 330 m