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 distance covered by the proton is 48.4 m
Explanation:
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is the surface charge density
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where the direction is toward the plane.
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where
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Substituting,
Now we can finally apply the following suvat equation for accelerated motion to find the distance travelled by the proton:
where
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is the initial velocity
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And solving for s,
Therefore, the closest answer is 48.4 m.
Learn more about accelerated motion:
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