Question
A golf ball (m = 35.4 g) is struck a blow that
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.
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.
Answers
Damon
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
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