Arrow- Finding the point of impact

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Homework Help Overview

The problem involves an archer shooting an arrow horizontally at a target, requiring calculations related to time of flight, impact point, and final velocity and angle of impact. The context is within kinematics, specifically projectile motion, with a focus on horizontal and vertical components of motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculations for time of flight and the vertical displacement of the arrow. There are attempts to clarify the use of equations related to projectile motion, particularly regarding the assumptions made about angles and acceleration due to gravity.

Discussion Status

Some participants have provided feedback on the calculations, indicating discrepancies in the results for the vertical displacement. There is acknowledgment of correct methods for certain parts, but also a recognition of errors in earlier attempts. The conversation reflects a collaborative effort to verify and correct the calculations without reaching a definitive conclusion.

Contextual Notes

Participants are working under the assumption that air resistance is negligible and are focusing on the effects of gravity alone. The discussion also highlights the potential for misunderstanding in applying the equations of motion.

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[SOLVED] Arrow- Finding the point of impact

Homework Statement



An archer shoots an arrow horizontally parallel to the ground at a height of 1.5 m,
above the ground, aimed directly at the centre of a target marked on a vertical wood panel situated 30 m from the archer. The arrow is released with an initial velocity of 75ms-1
Assuming air resistance to be negligible, determine (a) the time of flight of the arrow, (b) the point where the arrow hits the vertical wood panel, and (c) the final velocity and angle of impact of the arrow as it hits the vertical wood panel. Ignore the effects of any cross winds and arrow spin.

Homework Equations



x=[tex]vcos\theta[/tex] to work out time
[tex]v_y^2 = v_oy^2 + 2a_yy[/tex] for vertical velocity
[tex]v^2 = v_x^2 + v_y^2[/tex]
[tex]tan\theta[/tex] angle of impact

The Attempt at a Solution



I've already wokred out the time of flight of the arrow by dividing the horizontal range by the initial horizontal velocity and i got 0.4s. The part I'm struggling on is part b. Using the formula y=[tex]vsin\theta t - 1/2at^2[/tex] i got 0.98m. I don't think its right though, I asssumed a to be -9.81. For part (c) i worked out the vertical velocity , used pythagoras to get the resultant velocity and then used tan to find the angle. Could someone please tell if I'm doing the right thing for each part of the question. Although I'm pretty sure I've got it right for parts a&c
 
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For part b the arrow only has an acceleration due to gravity. I did not get the same answer as you did. For part c you have the correct method.
 
Stupid me. I worked it out wrong. the answer should be 0.785m(3sf). [tex]vsin\theta t - 1/2at^2[/tex] theta = 0 because its horizontal. a=-9.81 t=0.4s insert into equation and the answer should be 0.7848m.
 
thebestrc said:
Stupid me. I worked it out wrong. the answer should be 0.785m(3sf). [tex]vsin\theta t - 1/2at^2[/tex] theta = 0 because its horizontal. a=-9.81 t=0.4s insert into equation and the answer should be 0.7848m.

Very good. All sorted?
 
yes .thank you again. I don't know what's wrong with me today it's always the simple ones that put me off.
 

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