An archer fits a 0.31-kg arrow to a bowstring and then uses an average force of

Question

An archer fits a 0.31-kg arrow to a bowstring and then uses an average force of 130 N to draw the string back 1.1 m. If the archer releases the arrow at an angle of 45 degrees above the horizontal, from a height of 1.5 m, what is the maximum height the arrow will reach? Assume all the energy stored in the bowstring is transferred to the arrow and neglect air resistance.

answer options:

(A) 11 m

(B) 13 m

(C) 49 m

(D) 25 m

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Explanation / Answer

We use conservation of energy to find it's initial velocity

Elastic PE = 0.5Fx = 0.5( 130)(1.1) J

Kinetic energy = 0.5mv^2

0.5( 130)(1.1)= 0.5( 0.31)v^2

solving for v= 21.5 m/s aprx

Max height =( vsin)2/2g = ( 21.5 sin45)^2 / 19.6 =11.788 m apprx

adding to it the height = 1.5 m = 13 m aprx ( corrcet option is B

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