On the surface of the moon

Question

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!

Answer 1

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

$$a(t)=\frac{d^2x}{dt^2}=-5.3\frac{\text{ft}}{\text{s}^2}\;;$$

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

$$v(t)=\frac{dx}{dt}=-5.3t+C\;.$$

(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,

$$15=v(0)=-5.3\cdot0+C=C\;;$$

$C=15$, 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.

Answer 2

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\ $