What Is the Height of the Cliff If Both Balls Land Simultaneously?

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SUMMARY

The problem involves calculating the height of a cliff where two balls land simultaneously: one dropped from rest and another thrown downwards. The first ball is affected by a gravitational acceleration of 10 m/s² and falls for time t, while the second ball is thrown with an initial velocity of 20 m/s after a delay of 1.5 seconds. The key equation used is Xf = Xo + Vot - 1/2gt², leading to the conclusion that the height of the cliff can be determined by equating the distances traveled by both balls at the moment of impact.

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Homework Statement


A ball is dropped off a cliff on a planet with an exact gravity of 10m/s^2, 1.5 seconds later a lead ball is thrown straight down with an initial speed of 20 m/s If the two balls hit the base of the cliff at the same time, find the height of the cliff.


Homework Equations


Xf=Xo+Vot -1/2gt^2


The Attempt at a Solution

-d=-1/2gt^2
t^2=2d/g
t=\sqrt{2d/g}

-d=0-20(\sqrt{2d/g}-1.5)-1/2g(\sqrt{2d/g}-1.5)^2
 
Last edited:
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Wow, it took me a while to see that your last expression is d = Vi*t + .5at^2 with t replaced by (t-1.5). So that is your distance for the second ball at time t from when the first ball is dropped. To finish the job, you equate that distance to a similar expression for the first ball's distance.

If I may offer a more general tip, it would be to begin with
ball 1 distance = ball 2 distance
.5*a*t^2 = Vi*(t-1.5) + .5a(t-1.5)^2
so you have the big picture clearly stated, and then it is just detail work to finish it.
Also, you might take "down" to be positive in this problem to reduce the number of troublesome minus signs. I see it isn't really a quadratic equation - the t^2 term cancels a step or two further on.
 

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