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A marble is dropped a distance of 78.48 m. How long does it take to fall this far?

 Feb 14, 2019
 #1
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\(g = -9.8 ~m/s\\ h(t) = \dfrac{gt^2}{2} + v_0t+ h_0 = -4.9t^2 + 78.48\\ \text{at the end of it's fall the marble is at }h=0\\ 0=-4.9t^2 + 78.48\\ 4.9t^2 = 78.48\\ t^2 \approx 16.01632653\\ t \approx 4s\)

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 Feb 14, 2019
 #5
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Are there any other ways to figure this out? Any other equations to get the same answer.

Guest Feb 14, 2019
 #2
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To figure this out you need to have the speed which the ball is traveling.

V=s/t

Volocity equals speed over time.

 

Volocity is in meters per second

 Feb 14, 2019
 #3
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In this problem the velocity changes as the marble accelerates towards the ground.

Rom  Feb 14, 2019
 #4
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True. Your calculations are correct!

smiley

StarStrike  Feb 14, 2019

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