On the surface of the moon, acceleration due to gravity is approximately 5.3 feet per second squared. Suppose a baseball is thrown upward from a height of 6 feet with an initial velocity of 15 feet per second.

A) Dertimne the maximum height attained by the baseball.

I have to show all work, relevant calculus, and the appropriate position function.

What I am having trouble with is getting the problem started. Not sure what the position function would be and where I would go from there. I am really struggling in this calculus class and I need detailed help... Help please!

  • $\begingroup$ "Square feet per second" is not a unit of acceleration. $\endgroup$ Commented Jul 22, 2013 at 6:46
  • $\begingroup$ oops! i don't why i typed it like that, I meant feet per second squared. $\endgroup$
    – cschurman
    Commented Jul 22, 2013 at 6:52
  • 1
    $\begingroup$ A start, if this is to be done with calculus. Ground: $y=0$; up, positive; $t=0$ when ball is thrown; $y''=-4.3$. You can decide what $y(0)$ and $y'(0)$ should be. $\endgroup$ Commented Jul 22, 2013 at 6:57
  • $\begingroup$ i'm not really following what you're trying to say? $\endgroup$
    – cschurman
    Commented Jul 22, 2013 at 7:01
  • 1
    $\begingroup$ @Arjang It's called the SI system, and is only relevant in physics. With math we don't care about such earthly matters. We can do one just as well as the other. $\endgroup$
    – Arthur
    Commented Jul 22, 2013 at 9:26

2 Answers 2


We’ll measure distance $x$ in feet above the surface of the moon and time $t$ in seconds, with $0$ being the moment at which the ball is released. The ball’s velocity at time $t$ is then $v(t)=\frac{dx}{dt}$, the rate of change of position, and its acceleration is $a(t)=\frac{d^2x}{dt^2}$, the rate of change of velocity. You’re told that


it’s negative because it’s directed downwards. If you take the antiderivative of a second derivative, you get a first derivative:


(To avoid visual clutter, I’m omitting the units now.) In order to determine $C$, we must know the actual velocity at some specific moment. Fortunately, we do: at time $t=0$ the velocity is $15$ ft/s. Thus,


$C=15$, and we now know that $$\frac{dx}{dt}=v(t)=-5.3t+15\;.$$

At this point I’ll turn it over to you. You need use this to find a formula for $x(t)$, the height of the ball at time $t$; you can do that in much the way I just got a formula for $v(t)$ from the formula for $a(t)$. Once you have a formula for $x(t)$, use the usual calculus techniques to find its maximum.

  • $\begingroup$ this helps, but i am lost on how to find its maximum.... $\endgroup$
    – cschurman
    Commented Jul 22, 2013 at 7:43
  • $\begingroup$ @cschurman: Suppose that you were asked to find the maximum of the function $f(x)=x^2-2x-3$ on the interval $[0,3]$; what would you do? Once you have the function $x(t)$, you’ll find that your problem here is very similar to that hypothetical problem. $\endgroup$ Commented Jul 22, 2013 at 7:50
  • $\begingroup$ could you possibly help me and walk me through this problem in steps, I have fallen behind in class, and I am having a hard time understanding the maximization of a function, and I am not even sure how to maximize your hypothetical function. $\endgroup$
    – cschurman
    Commented Jul 22, 2013 at 9:24
  • $\begingroup$ @cschurman: For reasonably nice functions $f$, the maximum and minimum values of $f$ on a closed interval $[a,b]$ can only occur at $a$, at $b$, and at points $c\in[a,b]$ where $f\,'(c)=0$, the so-called critical points. To find the maximum, therefore, you find $f\,'(x)$, set it to $0$, and solve for the critical points. Then you compute the function values at those points and at $a$ and $b$ and pick out the largest. $\endgroup$ Commented Jul 22, 2013 at 9:31
  • $\begingroup$ I got this portion of the question figured out, the maximum height the baseball will reach is 33.45. there is more parts to the question, the next part reads "determine how long it takes the ball to hit the surface of the moon" i know i need to use the 2.83 seconds, and the 33.45 feet to find the answer for this part, just not quite sure what i need to do, could you help me with this? $\endgroup$
    – cschurman
    Commented Jul 22, 2013 at 9:46

Acceleration is given by $ a(t)=\frac{d^2s}{dt^2}=-5.3\ $, integrating acceleration function gives us the velocity function

$v(t)=\frac{ds}{dt}=-5.3t+ k\ $

at $ t=0 $   v(t) = $15 f/s $, therefore k= 15
the velocity function becomes $v(t)=-5.3t+ 15\ $.
Integrating the velocity function gives the displacement function
$s(t)=-5.3t^2+ 15t + k1\ $ at $t=0$, $k1= 6 $. hence
$s(t)=-5.3t^2+ 15t + 6\ $
the maximum height is given when $v(t)=\frac{ds}{dt}=0\ $
this is the same as $v(t)=\frac{ds}{dt}=-5.3t+ 15 = 0\ $, this gives $t=2.83 sec $ at maximum height
substituting is the displacement function, maximum height is:
$s(2.83)=-5.3(2.83)+ 15(2.83) + 6 = 33.45 feet\ $


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