Question

A ball is thrown directly upward from a height of 3 ft with an initial velocity of 20 ​ft/sec. The function ​s(t)=−16t2+20t+3 gives the height of the​ ball, in​ feet, t seconds after it has been thrown. Determine the time at which the ball reaches its maximum height and find the maximum height.

Answer 1

Answer:

$t=\frac{5}{8}s$ and $s(\frac{5}{8})=\frac{37}{4}ft$

Step-by-step explanation:

There are different approaches you can take to solve this problem. You can look at it from the geometric approach or the calculus approach. I will us the geometric approach.

When we analyze the given function for the height of the ball:

$s(t)=-16t^2+20t+3$

we can see that we are dealing with a quadratic function. This means that the graph will be an inverted parabola, just like the one in the attached picture.

so the idea is to find the vertex of the parabola to find the highest the ball gets to be and the time it gets to be at that height.

In order to do so, we can use the vertex formula, which looks like this:

$t=-\frac{b}{2a}$

this is use when you have a function with the form:

$f(x)=ax^2+bx+c$

in this case:

a=-16, b=20 and c=3, so the formula then turns to:

$t=-\frac{20}{2(-16)}$

which yields:

$t=\frac{5}{8} s$

which is our first answer. So we can then use this to find the maximum height of the ball, just by substituting t into the original function:

$s(\frac{5}{8})=-16(\frac{5}{8})^2+20(\frac{5}{8})+3$

which yields:

$s(\frac{5}{8})=\frac{37}{4} ft$

which is our second answer.