Solve Trajectory Motion Homework: Find t, Drop Apple to Pierce Arrow

In summary, an apple should be dropped after the arrow has been shot for 5.6 seconds in order for the apple to be pierced by the arrow.
  • #1
zellster87
15
0

Homework Statement

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?

Homework Equations


Vx = Vcos(theda), x= Vt



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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  • #2
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.
 
  • #3
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 don't understand how to manipulate the equation so my info fits. Thanks for your time.
 
  • #4
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.
 
  • #5
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.
 

Related to Solve Trajectory Motion Homework: Find t, Drop Apple to Pierce Arrow

1. How do I find the time (t) in a trajectory motion problem?

In order to find the time (t) in a trajectory motion problem, you will need to use the equation t = (v*sinθ)/g, where v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity. Plug in the known values and solve for t.

2. What is the acceleration due to gravity in a trajectory motion problem?

The acceleration due to gravity in a trajectory motion problem is a constant value of 9.8 m/s². This value is typically represented by the variable g in equations.

3. How can I determine the initial velocity (v) in a trajectory motion problem?

You can determine the initial velocity (v) in a trajectory motion problem by using the equation v = √(x*g)/(sin2θ), where x is the horizontal distance traveled and θ is the angle of launch. Plug in the known values and solve for v.

4. What is the angle of launch (θ) in a trajectory motion problem?

The angle of launch (θ) in a trajectory motion problem is the angle at which an object is launched or thrown. It is measured in degrees or radians and can be found using the equation θ = tan⁻¹(y/x), where x is the horizontal distance and y is the vertical distance. Plug in the known values and solve for θ.

5. Can I use the same equations for any trajectory motion problem?

Yes, the same equations can be used for any trajectory motion problem as long as the values for initial velocity, angle of launch, and acceleration due to gravity are known. However, keep in mind that these equations assume a few simplifying assumptions, such as no air resistance and a uniform gravitational field.

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