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Below I have a question that I tried to solve on a exam. I am curious as to the actual way to approach the question. What I did was set the equation equal to $0$ and get $t = -3$ then I plugged in $3$ into the derivative of the equation and I got $128$ ft/sec. But apparently I approached it wrong. What is the right way to approach the question and solve it for the questions below?

If a rock is thrown upward on Earth, with an initial velocity of 32 ft/sec from the top of a 48 ft building, we can model the height of the rock at time t by $s(t) = −16t^2+ 32t + 48$. After throwing the rock up, it will at some point be level again with the top of the 48 ft building.

What will the velocity be at this time?

Find the acceleration, why is it constant?

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    So in order to answer the secondpart of the question what steps do i follow?2012-06-24

3 Answers 3

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Recall that for constant acceleration only, the position of the object is given by:

$s = s_0 + v_ot + \frac {1}{2} t^2 \space \text{where v = velocity, s = position} \tag{1}$

Since our initial height and final height are both $48$ feet, we have:

$48 = 48 + 32t - 16t^2 \tag{2}$

Solving for the time to where the rock is at the same height it started at leads us to:

$0= -16t^2 + 32t \implies t = 0, 2 \space \text{sec} \tag{3} $

Obviously the rock is at the same height at $t = 0$, so at $t = 2$ sec is where it makes a parabola-like movement and comes back down to the original height of the $48$ foot building.

Taking the derivative of our position function, $s(t)$ and plugging in $t = 2$ sec leaves us:

$s'(t) = -32t + 32 \implies v = -32 \space \text{feet/sec} \tag{4} $

Taking the derivative again leads us to:

$s''(t) = -32 \space \text{feet per second per second} \tag{5}$

As expected, $s''(t)$ or rather $a$ for acceleration is indeed constant, since the ball is only accelerating due to gravity which is roughly $-32$ feet per second per second. Also, you could note that since acceleration is the second derivative of a position function and our position function was a quadratic, it follows that the acceleration function is constant since taking the derivative each time lowers the power. Further, as noted in equation (1), the formula only holds for constant acceleration, so it is a bit circular to see that since your given equation fits the model formula (1), the object undergoes constant acceleration.

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    Well $s'(t)$ or rather $v(t) = -32t + 32$. So, taking the derivative of the velocity function gives us the acceleration function. $s''(t) = v'(t) = -32.$ So, the acceleration is always a constant $-32$ feet per second per second.2012-06-25
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Note that we actually don't need to know $t$. By symmetry, the velocity when the rock reaches the "throw" level again must be the negative of the throw velocity.

Since the throw velocity is $32$, the answer to the first question must be $-32$ (feet per second).

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    So i$n$ order to a$n$swer the seco$n$dpart of the question what steps do i follow?2012-06-24
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When the rock is level again with the building, $s(t) = 48$. So solving for time,

$-16t^2 + 32t = 0$ gives $t=0,2$. So it appears that you just made a mistake in your calculation for $t$. Your method for finishing the question seems correct, just using $t=2$ instead.