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Hello everyone and thanks for taking the time to view my question. I've been stuck on the second part of this question now and can't seem to find the correct answer no matter what I do. The question is as follows:

A ball dropped from the top of a building has a height of $s = 144 - 16t^2$ meters after t seconds. How long does it take the ball to reach the ground? What is the ball's velocity at the moment of impact?

So for the first part, I found that it takes $3$ total seconds for the ball to reach the ground. However, when I try to find an answer for the second part I keep getting $48$ meters per second, when the correct answer is $96$ meters per second. Why is the correct answer double of the answer that I am getting?

Thank you for you help!

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The velocity is $$ v(t)=s^{\prime}(t)=-32t $$ Hence $v(3)=-96$.

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  • $\begingroup$ How did you get that equation? Since the building is 144 meters tall and it takes 3 seconds to cover that distance, wouldn't the speed of the ball be 48 meters per second? Thanks for the response. $\endgroup$ – Sorin Circa Apr 3 '16 at 23:13
  • $\begingroup$ I just differentiated $s(t)=144-16t^2$ using the power rule. The first term is a constant, so its derivative is zero. $\endgroup$ – carmichael561 Apr 3 '16 at 23:14
  • $\begingroup$ Also note that the velocity is a vector and has direction. The direction is negative, hence $v(3)=-96ms^{-1}$. Whereas the speed(a scalar) on impact is just $96ms^{-1}$. Sorry for being pedantic and it might seem like a minor detail but it is important. $\endgroup$ – Chris Daley Apr 3 '16 at 23:20
  • $\begingroup$ @Sorin 48 m/s is the average speed of the ball, not the speed at $t=3$. $\endgroup$ – Edward Jiang Apr 3 '16 at 23:21

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