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3 years ago

a hammer drops from a height of 8 meters. calculate the speed with which it hits the ground. show work

Physics

1 answer:

ioda3 years ago

3 0

Answer:

12.5 m/s

Explanation:

The motion of the hammer is a free fall motion, so a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2-u^2=2as$

Where, taking downward as positive direction, we have:

s = 8 m is the displacement of the hammer

u = 0 is the initial velocity (it is dropped from rest)

v is the final velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving the equation for v, we find the final velocity:

$v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(8)}=12.5 m/s$

So, the final speed is 12.5 m/s.

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$\sf{\pink{\underline{\underline{\blue{GIVEN:-}}}}}$

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The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

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Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

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$A=\sqrt{l^2-b^2}b$

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$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

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