The golf club is in contact with the golf ball for 1.8 ms and exerts a force of 1500 N on the golf ball. The mass of the golf ball is 0.045kg Calculate the velocity of the golf ball as it leaves the golf club. Velocity m/s Thank you :)

145m/s

Explanation:

Of course

Force x time = Impulse which is a change in momentum. You can then use this to find the velocity (as you know the ball's initial momentum is 0)

1500 x 1.8 x 10^-3 = 0.045v

v = (1500 x 1.8 x 10^-3)/0.045 = 60 ms^-1

Explanation:

The distance the ball travels is directly related to the force that is applied from the head of the golf club, the angle at which the force is applied, gravity, as well as air and wind resistance. The force of the club head on the ball causes the ball to compress and apply an equal amount of force on the club.

use ke=1/2mv^2 rearrange for V

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft