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3 years ago

I throw a ball off the edge of a 15.0m cliff. If I threw it horizontally at 8.0 m/s, how much time did it take to fall?

1 answer:

3 0

Just ignore the horizontal component

if you have a vertical displacement of 15m, 0ms^1 initial velocity, end velocity is ignored, we know the acceleration due to gravity as 9.81ms^2 so we can work out the time using SUVAT

S=15
U=0
V=?
A=9.81
T=?

S=UT + 0.5 AT^2

UT=0

therefore,

S=0.5AT^2

rearrange to:

T=SQR (2S/A)

T = 1.75 seconds

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Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

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3 years ago

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4 years ago

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