An archer shoots an arrow at a 75.0 m distant target, the bull\'s-eye of which i

Question

An archer shoots an arrow at a 75.0 m distant target, the bull's-eye of which is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull's-eye if its initial speed is 36.0 m/s? (Although neglected here, the atmosphere provides significant lift to real arrows.) Incorrect: Your answer is incorrect. ° (b) There is a large tree halfway between the archer and the target with an overhanging branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch? under over Correct: Your answer is correct.

Explanation / Answer

Here ,

distance , R = 75 m

initial speed , u = 36 m/s

for the range is given as

R = u^2 * sin(2 * theta)/g

75 = 36^2 * sin(2 * theta)/9.8

solving for theta

theta = 17.3 degree

the angle of release is 17.3 degree

b)

for half way

height = ( 36 * sin(17.3))^2/(2 * 9.8)

height = 5.83 m

it will go above the branch

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