Velocity Question & Acceleration

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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Need to set $-16t^2+32t+48=48$, because $48$ is the height when we pass the top of the building again. You can see (by putting $t=0$) that they are measuring distances from ground level. Definitely you must not put $-16t^2+32t+48$ equal to $0$. – André Nicolas Jun 23 '12 at 19:56
For the second, the acceleration is constant at -32 ft/sec^2. To see this, take two derivatives of $s(t)$ and this is what you get. This is an approximation, valid near the earth's surface, that gravity is the same strength at all places and directed downward. – Ross Millikan Jun 23 '12 at 23:48
So in order to answer the secondpart of the question what steps do i follow? – soniccool Jun 24 '12 at 23:45

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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You have obscured that between the first and second equations you redefined the origin without changing the variable. I think it would be clearer to say that s(0)=48 as that is the height of the building and when it is again level with the top of the building you have s(t)=48. You reach the same solution. – Ross Millikan Jun 23 '12 at 23:52
@Ross Yeah, I was a bit unclear. After a year of mechanics, you do these real quick, sorry. I edited the post in hopes of making it easier to understand. – Joe Jun 24 '12 at 1:36
So in order to answer the secondpart of the question what steps do i follow? – soniccool Jun 24 '12 at 23:46
@soniccool What do you mean the "second part"? The question regarding acceleration/why is it constant? – Joe Jun 25 '12 at 1:23
@JoeL. The question is What will the velocity be at this time? Find the acceleration, why is it constant? or is that the same thing that it is asking because i understand the velocity is -32ft/sec, but how do i go ahead to find the acceleration? – soniccool Jun 25 '12 at 1:25

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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Or one could invoke conservation of energy. – MJD Jun 23 '12 at 22:39
I don't think he has gotten to energy yet. Usually kinematics is taught before energy. – Joe Jun 23 '12 at 22:40
So in order to answer the secondpart of the question what steps do i follow? – soniccool Jun 24 '12 at 23:46

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.

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