A golfer strikes a 0.059-kg golf ball with a force of 290 N. If the ball moves with a velocity of 69 m/s,
1 answer:
Answer:
0.014s
Explanation:
Given parameters:
mass of golf ball = 0.059kg
force applied = 290N
velocity = 69m/s
initial velocity = 0m/s
Unknown:
Time of contact = ?
Solution;
We know that momentum is the quantity of motion of body possess;
Momentum = mass x velocity
Momentum = 0.059 x 69 = 4.1kgm/s
Also; impulse is the effect of the force acting on a body;
impulse = force x time = momentum
So;
Force x time = momentum
Time = =
= 0.014s
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Well, the density of the water is
so i believe that is what the question is asking for :)
so i believe that is what the question is asking for :)
Answer:
58.5 m
Explanation:
First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.
The initial vertical velocity of the ball is
where
u = 21.5 m/s is the initial speed
is the angle
Substituting,
The vertical position of the ball at time t is given by
where
h = 13.5 m is the initial heigth
is the acceleration of gravity (negative sign because it points downward)
The ball reaches the water when y = 0, so
Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.
The horizontal velocity of the ball is
And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows: