You shoot an arrow straight ahead; it leaves your bow moving parallel to the ground. You find it embedded in the ground 50m away from where you shot it. It makes an 82 degree angle with the level ground , reflecting the direction it was traveling at the time it hit. How fast was it going when it left your bow? (You can neglect air resistance)

Question

You shoot an arrow straight ahead; it leaves your bow moving parallel to the ground. You find it embedded in the ground 50m away from where you shot it. It makes an 82 degree angle with the level ground , reflecting the direction it was traveling at the time it hit. How fast was it going when it left your bow? (You can neglect air resistance)

So I figured this out...

yf=o
yf=yi+vyi(t)+1/2gt^2
vyi=0

vf= gt
so -yi=1/2gt^2

and xf=vx/t
vx remains constant

Fnet=Fg=-mg since gravity is the only thing acting on the arrow

<cos82, cos8,cos90> = unit vector of the arrow based on how it landed on the ground

unit vector = <vx,-gt,0>/vector magnitude

also
unit vector=<50m/t,-gt,0>vector magnitude

But then I got stuck, where should I go from here if I don't know the time or magnitude of the vector?

Answer 1

well, if fell a vertical distance of h in time t.

h=.5gt^2 or t=sqrth/2g
vv=g t or v=sqrt hg/2

but the horizontal velocity was 50/t

now looking at the angle at the end..
tan82=horizonal/vertical=50/t / gt
tan82=50/gt^2
solve for time t in the air.
then velocity initiual= 50/t

Answer 2

To solve the problem, you can break it down into two components: the horizontal motion and the vertical motion.

Let's start with the horizontal motion. Since the arrow was shot parallel to the ground and traveled 50 meters horizontally, we can use the formula for horizontal displacement:

xf = xi + vxi * t

In this case, xf is 50m, xi is 0m (where you shot the arrow from), and vxi is the horizontal component of the initial velocity. We are looking to find vxi.

Now let's look at the vertical motion. We can use the formula for vertical displacement, considering the arrow lands in the ground:

yf = yi + vyi * t + (1/2) * g * t^2

In this case, yf is 0m (since the arrow hits the ground), yi is also 0m (since you shot the arrow parallel to the ground), vyi is the vertical component of the initial velocity, and g is the acceleration due to gravity (-9.8 m/s^2).

Now, we have two equations and two unknowns: vxi and vyi. We can use trigonometry to relate these velocities to the initial velocity (v) and the launch angle (82 degrees).

First, let's find the magnitude of the initial velocity (v):

v = sqrt(vxi^2 + vyi^2)

Next, let's find the relationship between vxi, vyi, v, and the launch angle (82 degrees):

vxi = v * cos(82 degrees)
vyi = v * sin(82 degrees)

Now, substituting these expressions for vxi and vyi into the equations we derived earlier:

xf = v * cos(82 degrees) * t
0 = v * sin(82 degrees) * t + (1/2) * (-9.8 m/s^2) * t^2

Using the known value of xf (50m), we can solve the first equation for t:

t = xf / (v * cos(82 degrees))

Substituting this value of t back into the second equation, we can solve for v:

0 = v * sin(82 degrees) * (xf / (v * cos(82 degrees))) + (1/2) * (-9.8 m/s^2) * (xf / (v * cos(82 degrees)))^2

Now, you can simplify this equation and solve for v.

Once you find the value of v, that will be the magnitude of the initial velocity when the arrow left your bow.