Question

A ball is thrown across a playing field from a height of h = 3 ft above the ground at an angle of 45° to the horizontal at the speed of 20 ft/s. It can be deduced from physical principles that the path of the ball is modeled by the function

y = −32/(20^2)x^2 + x + 3

where x is the distance in feet that the ball has traveled horizontally.

(a) Find the maximum height attained by the ball. (Round your answer to three decimal places.)

(b) Find the horizontal distance the ball has traveled when it hits the ground. (Round your answer to one decimal place.)

Answer 1

(a) You can apply derivatives to find the maximum hegith, provided that f(x) has a possible maximum in x if f'(x)=0. So, first, derive y,
\[y′(x)=0\]Find the x that satisfy this equation. Once you get it, plug in this value in the original function, and you'll have the answer for (a).

Answer 2

Do you know how to apply derivatives?

Answer 3

no i need help i have no idea what to do

Answer 4

But do you know derivatives of a function?

Answer 5

no

Answer 6

Well, then, do you know what is a parabola?

Answer 7

yeah kinda not really

Answer 8

Well, let's do it with parabolas. The idea is find the vertex of the parabola that represents the function y(x). To find the vertex you can apply the usual formula,
\[h=ax^2+bx+c\]\[x_v=-\frac{b}{2a}\] In our case,
\[h=−32/(400)x^2 + x + 3\Rightarrow x_v=100/16=25/4\]Now you can find the maximum height plugin the value x into the function h(x),
\[h(x_v)=−32/(400)(25/4)^2 + (25/4) + 3=49/8\]That is the maximum height.

Answer 9

To find the horizontal distance, you only need to find where the hegith is 0.
\[y=0 = −32/(20^2)x^2 + x + 3\Rightarrow 0=−32x^2 +400 x +1200\]Using for,
\[ax^2+bx+c=0\]The quadratic formula,
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]You'll find the answer. Try it, and if you don't find the solution, tell me.

Answer 10

so I take −32x^2+400x+1200 and put it in the quadratic formula

Answer 11

Yes, this will give you the horizontal distance. Remember to select the positive solution.

Answer 12

oky im going to try it

Answer 13

Im stuck after putting it in the equation I am at -400 ± \[\sqrt{-2650}\]

Answer 14

\[x=\frac{-400\pm\sqrt{400^2+4\cdot32\cdot1200}}{2\cdot32}\]

Answer 15

It should give you x=15.

Answer 16

isnt the 4 and 32 suppose to be negative

Answer 17

well thanks for the help i really appreciate it !

Answer 18

Sorry,the 32 under the fraction is negative. ;)