Velocity of a ball thrown upwards
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A boy / girl (you choose!) throws a ball straight up with a velocity of 35.0 m/s.
a. How high is the ball when its velocity has slowed to 22.5 m/s?
b. How high is the ball when its velocity is -22.5 m/s?
c. What is its velocity when it is caught at the same height as when it was released?
d. What is the maximum height reached?
e. If the ball is thrown with the same velocity at an angle of 35o above the ground, what is its range?
f. What would it its maximum height be now?
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Solution Summary
The solution provides explanations to the mechanics of a ball thrown straight upwards and the force of gravity acting on it to change its velocity.
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boy / girl (you choose!) throws a ball straight up with a velocity of 35.0 m/s.
a. How high is the ball when its velocity has slowed to 22.5 m/s?
Solution:
a=-9.8m/s^2, V0=35m/s, V1=22.5m/s
V1=V0-at, we got:
t=(V0-V1)/a=1.276s
then, ...
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