A golf ball is struck at ground level. The speed of the golf ball as a function

Question

A golf ball is struck at ground level. The speed of the golf ball as a function of time is shown in the figure below, where t = 0 at the instant the ball is struck. The graph is marked in increments of 0.55 s along the time axis, and vmin = 18.76 m/s and vmax = 32.84 m/s. (Values in figure do not necessarily match values in problem).

(a) How far does the golf ball travel horizontally before returning to ground level?
m

(b) What is the maximum height above the ground level attained by the ball?
m

Explanation / Answer

Since vmin is 18.76 m/s, that is the x component of the velocity because the minimum is at the top of the path when there is no y velocity

The initial y velocity can then be found, since the x velocity does not change

vmax2 = vx2 + vy2

(32.84)2 = (18.76)2 + (vy)2

vy = 26.95 m/s

From that, we can find the angle of launch

tan = vy/vx = 26.95/18.76

= 55.16o

Part A)

The range formula is

R = vo2sin2/g

R = (32.84)2(sin 110.32)/9.8

R = 103.2 m (Range is the horizontal distance)

Part B)

The height is found using the y component

vf2 = vo2 + 2ad

(0) = (26.95)2 + (2)(9.8)(d)

d = 37.1 m

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