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A rock dropped from a high platform is moving at 24 m/s downward when it strikes the ground. Ignore air resistance. How fast was the rock moving when it had fallen only one-fourth of the distance to the ground? (Points : 3) 18.0 m/s 12.0 m/s 8.0 m/s 6.0 m/s
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since it is a uniformly accelerated motion, you could use one of the formulae for uniformly accelerated motion: \[v = u + at\] \[x = ut + \frac{ 1 }{ 2}at ^{2}\] \[v ^{2} = u ^{2} + 2ax\] x= displacement u= initial velocity v= final velocity a= acceleration t= time taken
does that help or i should explain further?
no perfect! thank you so much!!@kausarsalley!
you are welcome!
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