Question

# A golfer hits a 45 gram golf ball and it lands a distance away from the...

A golfer hits a 45 gram golf ball and it lands a distance away from the tee. The ball had an initial driving velocity of 45 m/s at 45 degrees. The ball lands 10 m on the ground up a slope from the hole. It has to roll down an in

a) Draw a picture of the problem labeling ALL the known variables.

b) What horizontal distance has the ball traveled from the golf tee when it hits the ground before rolling toward the hole?

c) What is the final magnitude of the velocity of the ball as it hits the ground?

d) Assuming all its velocity is transferred to its motion rolling down the hill to the hole, what is the kinetic energy of the ball right after it hits the ground and rolls?

e) What is the height of the hill that the ball must roll down?

f) What is the total energy of the ball when it reaches the hole?

Mass of golf ball, Mb = 0.045 kg

initial velocity, Vi = 45 m/s

angle , x = 45 degrees

vertical distance travelled by ball on the slope of hill = 10 m

a)

b) horizontal distance the ball traveled from the golf tee when it hits the ground before rolling toward the hole,

R = (Vi2 * sin (2x)) / g

=> R = 452 * sin(90) / 9.8

=> R = 206.63 m

c) magnitude of the velocity of the ball as it hits the ground

The initial and final (as the ball hits the ground) velocity, will be the same, however, the sign will change. The final velocity will be equal to the negative of the initial velocity. here, the magnitude of the velocity of the ball as it hits the ground, Vf = |Vi| => Vf = 45 m/s

d) the kinetic energy of the ball right after it hits the ground and rolls

K = (1/2)mV2

K = 0.5 * 0.045 * 45*45

K = 45.5625 J

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