Agolf ball (m = 46.0 g) is struck with a force that makes an angle of 44.4° with the horizontal. the ball lands 193 m away on a flat fairway. if the golf club and ball are in contact for 6.92 ms, what is the average force of impact? (neglect air resistance.)

Using the formula for bodies in projectile motion, we can calculate for the velocity v:

 d = (v^2) * (sin2theta) / g 

Substituting the given values:

193 m = (v^2) * (sin(2*44.4)) / (9.8 m/s^2)

v= 50.5 m/s

Calculating for the impulse I,

I = mf * vf – mi * vi

since mf = mi and vi = 0 (the golf ball starts form rest)

I = m * vf

I = 0.046 kg * 50.5 m/s

I = 2.323 kg m/s

Calculating for the force of impact:

I = F t

F = (2.323 kg m/s) / (6.92 x 10^-3 s)

F =335.7 kg m/s^2 = 335.7 N

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s