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3 years ago

10

Questions

1 answer:

astra-53 [7]3 years ago

8 0

Answer:

1968

Explanation:

2400*20.5*0.004

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A skier leaves the horizontal end of a ramp with a velocity of 31.0 m/s and lands 156.3 m from the base of a ramp how high is th

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<u>Answer:</u>

The height of ramp = 124.694 m

<u>Explanation:</u>

Using second equation of motion,

$s = ut + \frac{1}{2}at^2$

From the question,

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substituting values

$156.3 = 31\times t + 0$

t = $\frac{156.3}{31 }$

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Similary, for the case of landing

t = 5.042 s; initial velocity, u =0

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Substituting in $h = ut + \frac{1}{2}gt^2$

$h = 0 + \frac{1}{2} \times 9.81 \times (5.042)^2$

h = 124.694 m

So height of ramp = 124.694 m

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