Question

11 1/2 points| Previous Answers OsColPhysAP2016 3.4.P.029. y Notes Ask Your Teacher An archer shoots an arrow at a 74.0 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 hit the bull's-eye if its initial speed is 44.0 m/s? 17.3x (b) There is a large tree halfway between the archer and the target with an overhanging branch 3.42 m above the release height of the arrow. Will the arrow go over or under the branch? over under Additional Materials Reading

Answer 1

Given , Range of Projectile(R) = 74.0 m

Also R = [(u^2)*sin(2$\Theta$)]/g

where,

u = initial projectile velocity = 44.0 m/sec

$\Theta$ = angle of projection = ??

g = 9.81 m/sec^2

So,

sin(2$\Theta$) = R*g/(u^2)

$\Theta$ = 0.5*arcsin(R*g/(u^2))

$\Theta$ = 0.5*arcsin(74.0*9.81/(44.0)^2)

$\Theta$ = 11.0 deg

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