A golf ball (m = 35.4 g) is struck a blow that

Question

makes an angle of 61.8
◦ with the horizontal.
The drive lands 176 m away on a flat fairway.
The acceleration of gravity is 9.8 m/s
2
.
If the golf club and ball are in contact for
3.01 ms, what is the average force of impact?
Neglect air resistance.
Answer in units of N.

Answer 1

we must find the change of momentum during contact, in other words Vi and u
Vi = w sin 61.8
u = w cos 61.8
if w is the initial speed

176 = u t where t is total time in air

v = w sin 61.8 - g t
h = Hi + Vi t - 4.9 t^2
0 = 0 + Vi t - 4.9 t^2 = t(Vi - 4.9 t)
so it is at the ground at t = 0 and t = Vi/4.9

so
176 = u (Vi/4.9)
176 = u (w sin 61.8)/4.9
176 = w cos 61.8 (w sin 61.8) /4.9
862.4 = w^2 sin 61.8 cos 61.8
862.4 = w^2 sin (123.6) /2
2071 = w^2
w = 45.5 meters/second (whew !)
f = m w/3.01*10^-3
= .0354 * 45.5 *10^3/3.01
= 535 Newtons