A ball is thrown straight up into the air with a velocity of 12 m/s. Draw a motion diagram for the ball and then give as much qu
antitative information as you can about the motion
1 answer:
Answer:
find the diagram in the attachment.
Explanation:
Let vi = 12 m/s be the intial velocy when the ball is thrown, Δy be the displacement of the ball to a point where it starts returning down, g = 9.8 m/s^2 be the balls acceleration due to gravity.
considering the motion when the ball thrown straight up, we know that the ball will come to a stop and return downwards, so:
(vf)^2 = (vi)^2 + 2×g×Δy
vf = 0 m/s, at the highest point in the upward motion, then:
0 = (vi)^2 + 2×g×Δy
-(vi)^2 = 2×g×Δy
Δy = [-(vi)^2]/2×g
Δy = [-(-12)^2]/(2×9.8)
Δy = - 7.35 m
then from the highest point in the straight up motion, the ball will go back down and attain the speed of 12 m/s at the same level as it was first thrown
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The power in horsepower is 40.1 hp
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where
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is the change in height
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Answer:
The speed does it head toward the goal = 41.87
Explanation:
Mass = 0.107 kg
Initial velocity ( u ) = 0
Force (F) = 28 N
Time = 0.16 sec
From newton's second law, Force = mass × acceleration
⇒ F = m × a
⇒ 28 = 0.107 × a
⇒ a = 261.7 --------- (1)
This is the value of acceleration.
Final speed of the mass is calculated by the equation V = U + at
⇒ U = 0 because mass in in rest position at start.
⇒ V = a t
Put the values of acceleration and time in above formula we get
⇒ V = 261.7 × 0.16
⇒ V = 41.87
Therefore the speed does it head toward the goal = 41.87