# Thread: If you shoot an arrow, how do you find out how far it will go?

1. ## If you shoot an arrow, how do you find out how far it will go?

Theres the question:

A person shoots an arrow which leaves the bow at 4ft above the ground with an initial velocity of 88ft second at an elevation angle of 48*.

Will the arrow go over a 40 foot wall that is 200' away?

2. Um... I will use kinematics to solve this.

The vertical component of the arrow is

$88 \sin 48 = 65.4 ft/s$

The horizontal component of the arrow is $88 \cos 48 = 58.9 ft/s$

So, the time it takes to cover 200 ft is 200/58.9 = 3.40 s.

Now, use the formula $s = ut + \frac12 at^2$

with t = 3.40,
u = 65.4 ft/s
a = -9.8 m/s^2 (you will have to convert to ft/s^2)

and solve for s.

If s < (40 - 4), then the arrow won't pass. Otherwise, it will.

What did you get?

3. Hello, av8!

$\text{A person shoots an arrow which leaves the bow 4 ft above the ground}$
$\text{with an initial velocity of 88 ft/sec at an elevation angle of 48}^o.$

$\text{Will the arrow go over a 40-foot wall that is 200 feet away?}$

It certainly helps if you know the Trajectory Equations.

. . . . . $x \;=\;v_o\cos\theta\, t$

. . . . . $y \;=\;h_o + v_o\sin\theta\,t - 16t^2$

. . where: . $\begin{Bmatrix}
h_o &=& \text{initial height} \\
v_o &=& \text{initial velocity} \\
\theta &=& \text{initial angle} \end{Bmatrix}$

We are given: . $h_o = 4,\;v_o = 88,\;\theta = 48^o$

Our equations are:

. . $x \;=\;(88\cos48^o)t$

. . $y \;=\;4 + (88\sin48^o)t - 16t^2$

The question is: When $x = 200$, is $y \,\ge\,40$ ?

When $x = 200$, we have:
. . $(88\cos48^o)t \:=\:200 \quad\Rightarrow\quad t \:=\:\dfrac{200}{88\cos48^o} \:=\:3.396537613$

It takes about 3.4 seconds to travel 200 feet horizontally.

When $t = 3.4$, we have:

. . $y \;=\;4 + (88\sin48^o)(3.4) - 16(3.4^2) \;\approx\;41.4\text{ feet}$

Yes, the arrow will clear the wall.