calculus
A ball is thrown into the air with an initial velocity of $16 ft/s$. its height after $t$ seconds is given by $f(x) = 16t-4t^2$ . After how many seconds does the ball reach its maximum height?
I can't remember back whether this fell under the derivative section or the Limit section, therefore i am stuck
First you find the height after 2 seconds, we'll denote that as $y_2$. So using the fomula we have:
$$y_2 = 40 \cdot 2 - 16 \cdot 2^2 = 80 - 64 = 16$$
After 0.5 seconds, the time passed after the throw will be 2.5 seconds so for this height we have:
$y_{2.5} = 40\frac 52 - 16 \left(\frac 52\right)^2 =100 - 100 = 0$$
So after 2.5 second the height of the ball will be $0\text{ ft}$.
We know that the average velocity can be calulated using the following formula:
$$v = \frac{\Delta s}{\Delta t} = \frac{|y_1 - y_0|}{|t_1 - t_0|}$$
So after the substitution we have:
$$v = \frac{|0 - 16|}{|2.5 - 2|} = \frac{|-16|}{|0.5|} = \frac{16}{0.5} = 32$$
So the average speed in that period will be 32 $\frac{ft}{s}$
Now you can do the 0.1 second interval by yourself.
If you know the maximum height, the answer is really simple to find - we can directly work from there, knowing that the velocity at maximum height is $0 \frac m s$ because it is, essentially, not moving as it turns from upwards motion to downwards free-fall.
Note that there are two approaches to finding this solution - one, more basic and perhaps easier to understand (but longer and more complex), the kinematic approach, using basic ideas of velocity, acceleration, and displacement - and the other, although more advanced and, much simpler - the energy approach. I'll start with the basic solution using the kinematic approach.
We know that, for any object under constant acceleration, the velocity at any point in time can be modeled by
$$v = v_0 + at$$
As previously stated, because the ball falls starting from its maximum height, the initial velocity is $0$. Therefore, the equation becomes
$$v = at$$
where $v_0$ is the initial velocity, $a$ is acceleration, and $t$ is the time elapsed. For most free fall cases, $a = -g$ (where $g$ is gravitational acceleration) as gravitational acceleration is downwards. However, in this particular case, we already know the maximum height and only need to examine the part of the trajectory that is downwards - therefore, if we simply "rotate our axes," setting the direction of motion as the positive direction, we can set $a = g$ because gravity would be working in the same direction of our current motion. Therefore, we can modify the equation to become
$$v = gt$$
Now all that's left is finding the time elapsed to reach the ground. We know, first of all, that we can find the displacement of an object under constant acceleration by averaging the initial and final velocities and multiplying that by time, thus
$$\Delta x = \frac {t(v_0 + v)} 2$$
Again, here, we know $v_0 = 0$, so the equation becomes
$$\Delta x = \frac {t(v)} 2$$
From here, we can solve for $t$ and then substitute our solution for $t$ into the previous equation to solve for $v$.
\begin{align*} \Delta x &= \frac {t(v)} 2 \quad && \\ \frac {2\Delta x} v &= t \quad && \text{Multiply both sides by $\frac 2 v$} \\ t &= \frac {2\Delta x} v \quad && \text{Solve for $t$} \end{align*}
Then, we substitute that into the previous equation:
\begin{align*} v &= gt \quad && \text{Previous equation} \\ v &= g(\frac {2\Delta x} v) \quad && \text{Substitute for $t$} \\ v &= \frac {2g\Delta x} v \quad && \text{Simplify fractions} \\ v^2 &= 2g\Delta x \quad && \text{Multiply both sides by $v$} \\ v &= \sqrt{2g\Delta x} \quad && \text{Square root both sides} \end{align*}
Then, we can plug in everything we already know to solve for $v$. Because of personal preferences ;), I'll do it in SI units:
\begin{align*} v &= \sqrt{2 \cdot 9.81 \frac m {s^2} \cdot 40.23 m} \\ &\approx 28.04 \frac m s \\ &\approx 91.99 \frac {ft} s \text{ or } 92 \frac {ft} s \end{align*}
Assuming a system consisting of just the ground (gravitational force) and the ball, we can apply the Law of Conservation of Energy, which states that, in any closed system, the net energy (kinetic + potential) is constant. Here, we need only to find the energy at the two endpoints of motion and equate the two.
As previously stated, the velocity is 0 at the peak of motion - therefore, the only energy present would be potential energy. We know that gravitational potential energy is given by $mgh$, where $m$ is the mass of the ball, $g$ is the gravitational acceleration, and $h$ is the height. Thus
$$E = mgh$$
Here, when the ball is at the ground, the height $h$ is essentially $0$ (except that tiny (negligible) bit, as it isn't quite in contact with the ground yet), and all that needs to be measured is kinetic energy. We know that kinetic energy is given by $\frac 1 2 mv^2$, therefore
$$E = \frac 1 2 mv^2$$
Now, we can easily equate the energy representations at different points in the ball's trajectory at the two endpoints and solve for $v$.
\begin{align*} mgh &= \frac 1 2 mv^2 \quad && \text{Equate the two energy representations} \\ gh &= \frac 1 2 v^2 \quad && \text{Divide both sides by $m$} \\ 2gh &= v^2 \quad && \text{Multiply both sides by 2} \\ \sqrt{2gh} &= v \quad && \text{Square root both sides} \\ v &= \sqrt{2gh} \quad && \text{Solve for $v$} \end{align*}
Unsurprisingly, this yields the exact same result as it would with the kinematics approach - but is a much shorter and more concise solution - the only downside to this solution is that it requires a deeper understanding of Physics concepts of work and energy.
Best Answer
My advice is not to think of problems as being in "sections", but instead to think about what the problem is asking for.
That function is fairly well-behaved - no weird corners or jumps or holes.
So, where it reaches its maximum, it's going to be flat before falling back down, right?
HINT: If it's flat at its maximum what does that tell you about it?