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[SOLVED] Arrow- Finding the point of impact
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.
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
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
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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