How Long and Fast Does an Arrow Need to Travel to Clear a Tree?

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An archer aims to fire a 0.2kg arrow to just clear a 45m high tree located 60m away. The discussion focuses on calculating the time for the arrow to reach the required height and the necessary horizontal speed. Participants suggest breaking down the problem into horizontal and vertical components of motion, using relevant equations for projectile motion. The approach involves determining the initial vertical velocity needed to clear the tree and then solving a system of equations for time and speed. The consensus emphasizes the importance of working through the calculations methodically to find the solution.
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Homework Statement


An archer wants to fire a 0.2kg arrow so that it will just clear the top of a tree. The tree is 45m high and at a horizontal distance of 60m from the archer.

Calculate the time it will take the arrow to reach the required height and the horizontal speed it takes to reach the tree in this time.


Homework Equations


I'm not too sure...?


The Attempt at a Solution


I'm looking for some guidance to point me in the right direction rather than the answer. I WANT TO WORK IT OUT MYSELF! :D
 
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Welcome to PF, kidnappin.
As it stands, there are an infinite number of correct answers!
For each angle you choose, there is a speed that will work.
Could you have missed a given angle in the question?
We could solve it for the minimum initial speed. Any more information at all in the question?
 
I don't know, Delphi, I would assume it is implying the max height is the height of the tree as well. That's how I picture "just clear a tree". So you can find an initial vertical velocity by assuming that it clears the tree with 0 vertical velocity. Then you can use that to find the time and so on.
 
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That sounds sensible, Sethric!
In all these projectile motion problems, I like to start with two headings:
horizontal and vertical.
In the horizontal part, you have motion at constant speed so you write the formula for that, d = vt. Or x = Vx*t if you prefer.
In the vertical part, you have accelerated motion at a=g = -9.81 so a few formulas apply - usually V = Vy + at and y = Vy*t + 0.5*a*t² work very well. Here Vx and Vy are the horizontal and vertical components of the initial velocity V.

If you put in all the known numbers into the 3 equations, you will have 3 equations with three unknowns, t, Vx and Vy. Take t to be the time to maximum height, where Vy = 0, x = 60 and y = 45. Write those up and show us! After you get them right, the problem reduces to solving the system of 3 equations with 3 unknowns.
 
I would've used energy, but them again, I'm lazy.
 
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