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svp [43]
3 years ago
13

An object is thrown off of a cliff with a horizontal speed up 10 m/seconds. After 3 seconds the object hits the ground. Find the

height of the cliff and the total horizontal distance traveled by the object.
Physics
1 answer:
Margarita [4]3 years ago
4 0
Bro again the same type of sum..
well
it takes 3 second to hit the ground
the height will be
h = 5*3*3
h = 45m
horizontal distance would be
velocity*time
10*3 ,= 30m/s
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By Newton's second law,

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Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

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3) The material should be readily available (easily sourced)

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