A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the
1 answer:
Answer:
The average force on ball by the golf club is 340 N.
Explanation:
Given that,
Mass of the golf ball, m = 0.03 kg
Initial speed of the ball, u = 0
Final speed of the ball, v = 34 m/s
Time of contact,
We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :
So, the average force on ball by the golf club is 340 N.
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The initial height of the first body is given by:
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.