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3 years ago

8

A ball is thrown straight up from the ground with an unknown velocity. It reaches its highest point after 5.5 s. With what veloc

ity did it leave the ground?

1 answer:

Maslowich3 years ago

5 0

V = u + at
0 = u -9.8 x 5.5
u = 9.8 x 5.5 = 53.9 m/s

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Explanation:

Given:

Mass of Avi is, $m=40\ kg$

Spring constant is, $k=176,400\ N/m$

Compression in the spring is, $x=20\ cm=0.20\ m$

Let the maximum height reached be 'h' m.

Now, as the spring is compressed, there is elastic potential energy stored in the spring. This elastic potential energy is transferred to Avi in the form of gravitational potential energy.

So, by law of conservation of energy, decrease in elastic potential energy is equal to increase in gravitational potential energy.

Decrease in elastic potential energy is given as:

$EPE=\frac{1}{2}kx^2\\EPE=\frac{1}{2}\times 176400\times (0.20)^2\\EPE=88200\times 0.04=3528\ J$

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Now, increase in gravitational potential energy is equal decrease in elastic potential energy. Therefore,

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