(a) A person throws a rubber lacrosse ball (mass 0.250 kg) vertically upwards
with an initial speed of 9.81 m/s. The ball reaches the top of its trajectory and returns to
the person’s hand, travelling at 9.81 m/s in a downward direction. The positive y-direction
is taken to be vertically upward throughout this Exercise.
(i) Calculate the time the ball was in the air (from launch to the time at which it returned
to the person’s hand.)
(ii) Calculate the change of momentum py of the ball for the interval calculated in (i).
(iii) Based on the definition of impulse as a product of force and time interval, calculate the
impulse Jy delivered to the ball by the force of gravity (i.e., by the weight of the ball) in
the interval calculated in (i). Your answer should be the same as the change of momentum
found in part (ii).
(b) Now the person moves into a parking garage. The lacrosse ball is again thrown vertically
upward with the same initial speed. This time, the ball hits the horizontal concrete ceiling
of the garage 2.45 m above the launch point. The impulsive collision between ball and
ceiling lasts 0.0500 s. Immediately before and after the collision, the ball has the same
speed. However, after the collision the velocity of the ball is vertically downward, rather
than upward.
(i) Calculate the speed of the ball when it hit the ceiling, and find the change of momentum
of the ball during the 0.05 s collision.
(ii) Calculate the impulse (Jy)gravity delivered to the ball by the force of gravity during the
0.05 s collision.
(iii) You should find that the impulse due to gravity is almost negligible during the 0.05 s
collision. What is this impulse expressed as a fraction of the total impulse due to all forces
acting on the ball in the collision?
(iv) Use the impulse-momentum theorem to calculate the impulse (Jy)ceiling delivered to the
ball by the contact force exerted on it by the concrete ceiling.
(v) Calculate the average force exerted by the ceiling on the ball during the 0.05 s collision.
2006-11-18 07:21:50 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
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2006-11-18 07:22:57 · answer #1 · answered by god knows and sees else Yahoo 6 · 0⤊ 0⤋
The answer is A When you through the ball up in the air it has a velocity/speed of X. Gravity will slow the ball until it has no more upward velocity/speed. Gravity will start pulling the ball back down again, increasing in velocity/speed until you catch it again. Gravity is a constant. The distance between your hands and the point at which the ball stopped is the same for both the upward and downward trip. It takes the same amount of time to slow the ball as it does to speed it back up while falling. SR
2016-05-22 01:17:36 · answer #2 · answered by Anonymous · 0⤊ 0⤋