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Trajectory Motion

by zellster87
Tags: motion, trajectory
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zellster87
#1
Jan29-08, 09:28 PM
P: 15
1. The problem statement, all variables and given/known data Hi, I have been working on this problem for a while, and cannot figure it out. Heres the question.

An arrow is shot at an angle of theta = 45 degrees above the horizontal. The arrow hits a tree a horizontal distance D = 220m away, at the same height above the ground as it was shot. Use g = 9.8m/s^2 for the magnitude of the acceleration due to gravity

a.)Find t, the time that the arrow spends in the air.
b.)Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree?
2. Relevant equations
Vx = Vcos(theda), x= Vt



3. The attempt at a solution

a.) a = 2x / t^2, 9.8 = 2(220cos45) / t^2, t = 5.6

5.6 is not the right answer. Thanks for your time.
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rocomath
#2
Jan29-08, 09:48 PM
rocomath's Avatar
P: 1,755
So we'll need to equations ...

[tex]x=(v_0\cos\theta)t[/tex]

Solve for initial velocity in the x, then plug that in [tex]y=(v_0\sin\theta)t-\frac{1}{2}gt^2[/tex] and solve for t.
zellster87
#3
Jan29-08, 10:02 PM
P: 15
Thanks for your response,but I still can't get it. With the equation x = Vcos(theda)t there is 2 unknowns, time and velocity. I dont understand how to manipulate the equation so my info fits. Thanks for your time.

rocomath
#4
Jan29-08, 10:32 PM
rocomath's Avatar
P: 1,755
Trajectory Motion

Solving for velocity in the x ...

[tex]v_0=\frac{x}{t\cos\theta}[/tex]

Plugging into y ...

[tex]y=\frac{x}{t\cos\theta}(\sin\theta)t-\frac{1}{2}gt^2[/tex]

In your problem ... "at the same height above the ground as it was shot" so y=0.

Solve for t.
zellster87
#5
Jan29-08, 11:02 PM
P: 15
Ah, it finally makes sense. Thanks for your reply, it really helped.

On part b i left out part of the question on accident. It said that "Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree." How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree?

I know that if there was no angle,then these would hit the ground at the same time, regardless of the velocity. But since there is an initial angle, how do I take that into account for my problem. Thanks again for your help.


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