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2 years ago

A child throws a ball up with an initial velocity of 4

Physics

1 answer:

raketka [301]2 years ago

5 0

The height of the ball after the given time of motion is 3.275 m.

The given parameters;

initial velocity of the ball, u = 4 m/s
height of fall of the ball, h = 2.5 m
time of motion, t = 0.5 s

The distance traveled by the ball is calculated as follows;

$h =h_0 + ut - \frac{1}{2}gt^2 \\\\h = 2.5 + 4(0.5) - (0.5\times 9.8\times 0.5^2)\\\\h = 3.275 \ m$

Thus, the height of the ball after the given time of motion is 3.275 m.

Learn more here:brainly.com/question/21180604

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Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

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Explanation:

Set up force equation

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$\frac{mv^{2} }{R}=W+F_{B}\\ F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N$

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