Register to reply

Basic projectile motion problem

by nick727kcin
Tags: basic, motion, projectile
Share this thread:
nick727kcin
#1
Feb11-06, 12:59 PM
P: 160
...

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....
Phys.Org News Partner Science news on Phys.org
Fungus deadly to AIDS patients found to grow on trees
Canola genome sequence reveals evolutionary 'love triangle'
Scientists uncover clues to role of magnetism in iron-based superconductors
nick727kcin
#2
Feb11-06, 01:04 PM
P: 160
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
#3
Feb11-06, 01:44 PM
PF Gold
daniel_i_l's Avatar
P: 867
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).

nick727kcin
#4
Feb12-06, 10:06 AM
P: 160
Basic projectile motion problem

ok so this is the real time:



time=.8208s
dx= 3.30 m


dx= vt

v= 76.85 m/s but this is wrong. ahhh this is killll
arildno
#5
Feb12-06, 10:13 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
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:
[tex]63=Vt[/tex]
Vertical distance traveled:
[tex]H=\frac{gt^{2}}{2}[/tex]
Ratio of velocities at time of impact:
[tex]\frac{V_{y}}{V_{x}}=-\tan(3), V_{y}=-gt, V_{x}=V[/tex]

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.
nick727kcin
#6
Feb12-06, 10:22 AM
P: 160
Quote Quote by 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:
[tex]63=Vt[/tex]
Vertical distance traveled:
[tex]H=\frac{gt^{2}}{2}[/tex]
Ratio of velocities at time of impact:
[tex]\frac{V_{y}}{V_{x}}=-\tan(3), V_{y}=-gt, V_{x}=V[/tex]

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 alot man
arildno
#7
Feb12-06, 10:31 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
Eeh, not too sure what you've done here!

From the 3rd equation, we get:

[tex]t=V\frac{\tan(3)}{g}[/tex]
and hence, from the first:
[tex]V=\sqrt{\frac{63g}{\tan(3)}}\approx\sqrt{\frac{60*63*g}{\pi}}\approx{60 }\sqrt{3}[/tex]
nick727kcin
#8
Feb12-06, 10:33 AM
P: 160
Quote Quote by arildno
Eeh, not too sure what you've done here!

From the 3rd equation, we get:

[tex]t=V\frac{\tan(3)}{g}[/tex]
and hence, from the first:
[tex]V=\sqrt{\frac{63g}{\tan(3)}}\approx\sqrt{\frac{60*63*g}{\pi}}\approx{60 }\sqrt{3}[/tex]
ooo i get it. you substituted vy=-gt into the third equation. thanks so much man:
arildno
#9
Feb12-06, 10:52 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
Quote Quote by nick727kcin
Well, I'm not going after your home address, but I am flattered.


Register to reply

Related Discussions
Projectile motion problem Introductory Physics Homework 2
Basic Projectile Motion question. Introductory Physics Homework 9
Projectile motion problem Introductory Physics Homework 15
Projectile Motion + Linear Motion = Problem Introductory Physics Homework 3
Uniform Circualr Motion with Projectile Motion problem (Extremely confusing): Introductory Physics Homework 4