An archer shoots an arrow at an 84.0 m distant target; the bull's-eye of the target is at same height as the release height of the arrow. (a) At what angle in degrees must the arrow be released to hi

asked by @student85195 • about 1 year ago • Physics • 5 pts

t the bull's-eye if its initial speed is 36.0 m/s?

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1 answer

V0 is 36.0 m/2, delta x is 84.0m

RANGE=84 m ==> (V0^2)sin2θ/g = 84 m

sin2θ=84 m (g/(v0)^2) 2θ =arcsin(84 m[(9.81m/s^2)/(36 m/s)^2] 2θ=39.48 deg θ =19.74 deg

answered by @shungoo1 • about 1 year ago
  • Correct! Juan commented about 1 year ago