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

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The discussion revolves around solving a trajectory motion problem involving an arrow shot at a 45-degree angle and an apple dropped from a height of 6.0 meters. The user initially struggles to find the time the arrow spends in the air and attempts to use the equations of motion but finds inconsistencies in their calculations. After some guidance, they begin to understand how to manipulate the equations to solve for time and velocity correctly. For part b, the user seeks clarification on how to account for the initial angle of the arrow when determining the timing for dropping the apple to ensure it is pierced by the arrow. The conversation highlights the importance of correctly applying kinematic equations in projectile motion problems.
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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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So we'll need to equations ...

x=(v_0\cos\theta)t

Solve for initial velocity in the x, then plug that in y=(v_0\sin\theta)t-\frac{1}{2}gt^2 and solve for t.
 
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.
 
Solving for velocity in the x ...

v_0=\frac{x}{t\cos\theta}

Plugging into y ...

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

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

Solve for t.
 
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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