Question

An archer releases an arrow from a shoulder height of 1.39 m. When
the arrow hits the target 18 m away, it hits point A. When the target is
removed, the arrow lands 45 m away. Find the maximum height of the
arrow along its parabolic path.

Ive seen this question on this website but nobody ever came up with a clear answer ive been trying to figure it out for hours can somebody help??

Answer 1

Anyone?? about to throw the towel in on this one

Answer 2

Let's not throw the towel JUST YET haha

Answer 3

I would say that you know the change in the y-value is -1.39m over 45m right?

Answer 4

So then we know that...\(\Delta x=45m\) and also we know that if we have an initial speed at an angle, \(v_i\), it can be broken up into x, and y components:

With projectile motion, the x-component remains constant so then we know that:
\[v_{ix}=v_i\cos\theta=\frac{\Delta x}{\Delta t}\]
So then, therefore, we know that for a y-displacement of \(\Delta y\), a time of \(\Delta t\), an acceleration of \(a\), and an initial speed of \(v_i\), we have:
\[\eqalign{
\Delta y=\ v_{iy}\Delta t+\frac{a}{2}\Delta t^2
}\]

Answer 5

would you like to see what i have so far? this is confusing and very frustrating

Answer 6

Sure! Let's see! I admit, it is NOT a very easy question haha

Answer 7

y=f(x)=ax^2+bx+c
f(0)=1.39
f(18)=h
f(45)=0
a(0)^2+b(0)+c=1.39
a(18)^2+b(18)+c=h
a(45)^2+b(45)+c=0
324a+18b=h-1.39
2025a+45b=-1.39

Answer 8

So you have three equations:
\[\eqalign{
&a(0)^2+b(0)+c=1.39\to c=1.39\\
&a(18)^2+b(18)+c=h\to324a+18b+1.39=h \\
&a(45)^2+b(45)+c=0\to2024a+18b+1.39=0 \\
&\\
&\\
&\\
&324a+18b+1.39=h\to a=\frac{h-1.39-18b}{324}\\
&2024a+45b+1.39=0\to a=-\frac{(45b+1.39)}{2024}\\
&\\&\\&\\
&\frac{h-1.39-18b}{324}=-\frac{(45b+1.39)}{2024} \\
&2024(h-1.39-18b)=-324(45b+1.39) \\
&2024h-2813.36-36432b=-14580b-450.36 \\
&\\
&h=\frac{21852b+2363}{2024} \\
}\]

Answer 9

So that is the max height of its trajectory?

Answer 10

well, as a function of the \(b\) value, yes, but otherwise, not a complete number

Answer 11

Err I got to go! I'll take a look at this in more depth later!

Answer 12

someone help