Arrow- Finding the point of impact

  • Thread starter Thread starter thebestrc
  • Start date Start date
  • Tags Tags
    Impact Point
AI Thread Summary
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.
thebestrc
Messages
10
Reaction score
0
[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
 
Last edited:
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top