An archer shoots an arrow at a 77.0 m distant target, the bull\'s-eye of which i

Question

An archer shoots an arrow at a 77.0 m distant target, the bull's-eye of which is at same 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 is 38.0 m/s? (Although neglected here, the atmosphere provides significant lift to real arrows.) degree (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? over under A basketball player is running at 4.80 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise 0.650 meters above the floor? m/s (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket? m

Explanation / Answer

(5) ans

Given that

Range R=77 m

initial speed u=38 m/s

now we find the direction

R=u^2sin2(theta)/g

77=38^2sin2(theta)/9.8

sin(2*theta)=0.523

2**theta=31.5

theta=15.75 degree

now we find the maximum height of the arrow

Hmax=38^2*sin^2(15.75)/2*9..8=5.43

so the arrow moves over the tree

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(6) ans

Given that

initial velocity u=4.8 m/s

height h=0.65 m

now we find the vertical velocity

v^2-4.8^2=-2*9.8*0.65

v^2=10.3

velocity v=3.21 m/s

now we find the range

range R=4.8^2/9.8=2.4 m

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