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

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AI Thread Summary
To find the height of the cliff, the problem involves two balls: one dropped and one thrown downwards. The first ball falls under gravity, while the second ball is thrown with an initial speed of 20 m/s, 1.5 seconds later. The key equations involve calculating the distances traveled by both balls and equating them, taking into account the time difference. A suggestion is made to simplify the problem by considering downward motion as positive to avoid negative signs. The final solution requires careful manipulation of the equations to determine the height of the cliff.
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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
 
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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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