Basic arrow projectile motion problem

[pureouchies4717]

...

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 63.0 m away, making a 3.00 degree angle with the ground. how fast was the arrow shot?

look two posts down...

 

[pureouchies4717]

using another equation...

d= vit +.5at^2
63= vi(.616)+.5(-187)(.616)^2
vi= 678.23m/s

which is also wrong...

in case a wasnt supposed to be negative:

63=vi(.616)+.5(187)(.616^2)
vi= -473.69 m/s

which is also wrong

 

[daniel_i_l]

I'm not sure exactly what you did but for one thing, there is no acceleration in the x direction, only downwards cause of energy.
So the Vox = Vix. Cause of that you know that the x component of the original speed times the time equals the distance (63).
Also, the y component of the original speed Viy goes with the equation:
H = Viy*t - 0.5g*t^2. The H you have (dy, but I got 3.3) so now you have two equations with two unknowns (t and Vi).

 

[arildno]

First, be absolutely clear on what are your principal unknowns here!

Those are
a) the initial velocity V in the x-direction
and
b) The height H from which the archer shot the arrow parallell to the ground (i.e, the initial vertical velocity is, indeed 0, as you thought).

c) The time t it took the arrow to hit the ground.
Let's set up a few equations relating V,t,H:

Horizontal distance traveled:
$$63=Vt$$
Vertical distance traveled:
$$H=\frac{gt^{2}}{2}$$
Ratio of velocities at time of impact:
$$\frac{V_{y}}{V_{x}}=-\tan(3), V_{y}=-gt, V_{x}=V$$

The last equation expresses that the arrow at the time of impact follows is parallell to the tangent line of the parabolic arc at that point.

You only need the first and third equation to answer your question.

 

[pureouchies4717]

First, be absolutely clear on what are your principal unknowns here!
Those are
a) the initial velocity V in the x-direction
and
b) The height H from which the archer shot the arrow parallell to the ground (i.e, the initial vertical velocity is, indeed 0, as you thought).

c) The time t it took the arrow to hit the ground.
Let's set up a few equations relating V,t,H:

Horizontal distance traveled:
$$63=Vt$$
Vertical distance traveled:
$$H=\frac{gt^{2}}{2}$$
Ratio of velocities at time of impact:
$$\frac{V_{y}}{V_{x}}=-\tan(3), V_{y}=-gt, V_{x}=V$$

The last equation expresses that the arrow at the time of impact follows is parallell to the tangent line of the parabolic arc at that point.

You only need the first and third equation to answer your question.

thanks for your help. so...
t=.8209s
Vy= -8.0444m/s

final result:

Vy/[-tan(3)]=Vx
Vx=[-gt]/[tan(3)]
vx=153.5 m/s

its still saying its wrong, but thanks a lot man

 

[arildno]

Eeh, not too sure what you've done here!

From the 3rd equation, we get:

$$t=V\frac{\tan(3)}{g}$$
and hence, from the first:
$$V=\sqrt{\frac{63g}{\tan(3)}}\approx\sqrt{\frac{60*63*g}{\pi}}\approx{60}\sqrt{3}$$

 

[pureouchies4717]

Eeh, not too sure what you've done here!

From the 3rd equation, we get:

$$t=V\frac{\tan(3)}{g}$$
and hence, from the first:
$$V=\sqrt{\frac{63g}{\tan(3)}}\approx\sqrt{\frac{60*63*g}{\pi}}\approx{60}\sqrt{3}$$

ooo i get it. you substituted vy=-gt into the third equation. thanks so much man: :!)

 

[arildno]

:!)

Well, I'm not going after your home address, but I am flattered.