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?
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?
Um... I will use kinematics to solve this.
The vertical component of the arrow is
$\displaystyle 88 \sin 48 = 65.4 ft/s$
The horizontal component of the arrow is $\displaystyle 88 \cos 48 = 58.9 ft/s$
So, the time it takes to cover 200 ft is 200/58.9 = 3.40 s.
Now, use the formula $\displaystyle s = ut + \frac12 at^2$
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?
Hello, av8!
$\displaystyle \text{A person shoots an arrow which leaves the bow 4 ft above the ground}$
$\displaystyle \text{with an initial velocity of 88 ft/sec at an elevation angle of 48}^o.$
$\displaystyle \text{Will the arrow go over a 40-foot wall that is 200 feet away?}$
It certainly helps if you know the Trajectory Equations.
. . . . . $\displaystyle x \;=\;v_o\cos\theta\, t $
. . . . . $\displaystyle y \;=\;h_o + v_o\sin\theta\,t - 16t^2 $
. . where: .$\displaystyle \begin{Bmatrix}
h_o &=& \text{initial height} \\
v_o &=& \text{initial velocity} \\
\theta &=& \text{initial angle} \end{Bmatrix}$
We are given: .$\displaystyle h_o = 4,\;v_o = 88,\;\theta = 48^o$
Our equations are:
. . $\displaystyle x \;=\;(88\cos48^o)t $
. . $\displaystyle y \;=\;4 + (88\sin48^o)t - 16t^2$
The question is: When $\displaystyle x = 200$, is $\displaystyle y \,\ge\,40$ ?
When $\displaystyle x = 200$, we have:
. . $\displaystyle (88\cos48^o)t \:=\:200 \quad\Rightarrow\quad t \:=\:\dfrac{200}{88\cos48^o} \:=\:3.396537613$
It takes about 3.4 seconds to travel 200 feet horizontally.
When $\displaystyle t = 3.4$, we have:
. . $\displaystyle y \;=\;4 + (88\sin48^o)(3.4) - 16(3.4^2) \;\approx\;41.4\text{ feet}$
Yes, the arrow will clear the wall.