Show work for each part of the question For constant acceleration a in 1 dimensi

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Show work for each part of the question

For constant acceleration a in 1 dimension: v=vo + at x-Xo Vo t +1/2afx-Xo-1/2 (Vo+ v)t x-xo=vt-1/2 ap +2 a(x -Xo) sin (26%) y(x) = (tana,)x-2moos_ R= Projectile motion: 1. An archer (on Earth) releases an arrow with vo 51 m/s at an angle of 16°, and hits the b target. If y-0 is the ground, the bullseye is at height y-1.5 m, as is the arrow at moment a) [6 pts] Give values for each of Xo. Yo. Vox. Voy, ay, and a,. If there is not enough in ngle of 16, and hits the bullseye on a of release (t-0). b) [6 pts] How far away is the target from the archer? c) [6 pts] How long, in seconds, is the arrow in the air before hitting the bullseye? d) [2 pts] The archer makes a second shot with different Vo and 6, and a spectator observes that y=1.1 m at t-2.75 s. Assume that 0>0. Will the arrow hit the bullseye? (Yes/No/Not Enough Info) Explain. Hint: You can answer this question without calculation.

Explanation / Answer

a) Vox= 51 cos 16= 49 m/s , Voy = 51 sin 16,= 14.06 m/s apprx

ay = -9.8 m/s^2 , ax = 0 m/s^2

x0 = ? ( not really sure what has to be calculated), y0 ( not really sure what has to be calculated)

b)h = ut - 1/2 gt^2

1.5 = 14.06 (t) -1/2 (9.8) t^2

4.9 t^2 - 14.06 t + 1.5=0

t = 14.06 +- 12.97/ 9.8 = 2.758, 0.111

there are two times, one while the arrow is rising and second while arrow is descending

Distance from archer will be out of thees two=  49 m/s (2.758) = 135.142 m , 49 (0.111)= 5.45 m

c)either 2.758 or 0.111

d ) height of arrow with respect tp ground at 2.75 seconds should have been 1.5 m to hit the bull's eye

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