A basketball is shot at 14.0 m/s at a 65.0 degree angle. What is the magnitude only (no direction) of the velocity of the ball 2
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Answer:
1.0752 kgm/s
Explanation:
Considering when the drop was dropped from rest from a height,
mass of the ball, m = 0.120 kg
height, h = - 1.25 m
the initial velocity, u = 0 m/s
the acceleration due to gravity, g = - 9.8 m/s²
From equation of motion
Substituting the values,
V = ± 4.95 m/s
V = - 4.95 m/s
Since the ball is moving downward, the final velocity of the ball when it hits the floor is V = - 4.95 m/s
Considering when the ball rebounds from the floor,
assume the mass of the ball still remain, m = 0.120 kg
height, h = 0.820 m
the final velocity, v = 0 m/s
the acceleration due to gravity, g = - 9.8 m/s²
From equation of motion
Substituting the values,
U = ± 4.01 m/s
U = + 4.01 m/s
Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is U = + 4.01 m/s
From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.
⇒ Impulse = Change in momentum
To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,
∴ F.t = 0.120 kg(4.01 m/s - (-4.95 m/s) )
F.t = 0.120 kg(4.01 m/s + 4.95 m/s) )
F.t = 0.120 kg × 8.96 m/s
Impulse = 1.0752 kgm/s
The impulse given to the ball by the floor is 1.0752 kgm/s