Arrow- Finding the point of impact

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An archer shoots an arrow horizontally at a height of 1.5 m with an initial velocity of 75 m/s towards a target 30 m away. The time of flight is calculated to be 0.4 seconds. The vertical drop of the arrow, determined using the formula for vertical motion, is corrected to approximately 0.785 m. The final velocity and angle of impact are also calculated correctly using the appropriate physics equations. Overall, the problem was resolved with the correct application of formulas despite initial confusion.
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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=vcos\theta to work out time
v_y^2 = v_oy^2 + 2a_yy for vertical velocity
v^2 = v_x^2 + v_y^2
tan\theta 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=vsin\theta t - 1/2at^2 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). vsin\theta t - 1/2at^2 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). vsin\theta t - 1/2at^2 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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