A man stands 5 m above the fairway and drives a golf ball with an initial veloci

Question

A man stands 5 m above the fairway and drives a golf ball with an initial velocity of v = <0,20,49> m/s. The man wishes to impart slice to the golf ball, which is modeled by an acceleration of 1.2 m/s^2 in the x direction. Thus, the acceleration function is given by: a(t) = <1.2, 0, -9.8>.

Assuming r(0) = <0, 0, 5>, determine:

a) The velocity and position functions.

b) The maximum height of the golf ball.

c) The rage of the shot; that is, the distance between where the ball lands and <0, 0, 0>.

NOTE: Please do not answer unless you are absolutely sure!!!

Explanation / Answer

(a)
Vx = 0
Vy = 20 m/s
Vz = 49 m/s

ax = 1.2 m/s^2
ay = 0
axz = -9.8 m/^2

V(t) = Vo + a(t) * t
V(t) = <0,20,49> + t *  <1.2, 0, -9.8>
V(t) = <1.2t , 20 , 49 - 9.8t >

S(t) = r(o) + vo* t + 1/2 * a(t) * t^2
S(t) = <0, 0, 5> + <0,20,49>  *t + 1/2 * <1.2, 0, -9.8> * t^2
S(t) = <0.6 t^2 , 20 t , 5 + 49*t - 4.9 *t^2>

Velocity Function, V(t) = <1.2t , 20 , 49 - 9.8t >
Position Function, S(t) = <0.6 t^2 , 20 t , 5 + 49*t - 4.9 *t^2>

(b)
Max Height reached by ball, H

v^2 = u^2 - 2*g*h
h = 49^2 / (2*9.8)
h = 122.5 m

H = h + 5 m
H = 122.5 + 5
H = 127.5 m
Maximum height of the golf ball, H = 127.5 m

(c)
Time taken by ball to reach back ground
Time to reach max height =
v = u - a*t
t = u/a
t = 49/9.8
t = 5 s

time taken to reach back ground,
s = u*t +1/2*a*t^2
127.5 = 0 + 1/2 * 9.8 * t^2
t = 5.1 s

Total time taken to reach ground = 5 + 5.1 = 10.1 s

Position in 10.1 s

S(t) = <0.6 t^2 , 20 t , 5 + 49*t - 4.9 *t^2>
S(t) = <0.6 *10.1^2 , 20*10.1 , 5 + 49*10.1 - 4.9 *10.1^2>
S(t) = <61.2, 202 , 0>

Range of Short, = sqrt(61.2^2 + 202^2)
Range of Short, = 211.1 m

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