# An archer places a 0.15 kg arrow on a bowstring. Then he exerts an average force of 100N to draw the string back 0.45m. Assume friction is negligible. What speed does the bow give the arrow and if the arrow is shot vertically upwards into the air, how high will it go? The archer places a 0.15 kg arrow on the bowstring and draws it back 0.45 m with an average force of 100N. The work done by the archer is 100*0.45 = 450 J. This is the potential energy stored that is converted to kinetic energy when the arrow is released.

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The archer places a 0.15 kg arrow on the bowstring and draws it back 0.45 m with an average force of 100N. The work done by the archer is 100*0.45 = 450 J. This is the potential energy stored that is converted to kinetic energy when the arrow is released.

When the arrow is released the kinetic energy of the arrow is equal to the potential energy stored. The kinetic energy of  body with mass m traveling with a velocity v is `(1/2)*m*v^2` . If the speed of the arrow is v, `(1/2)*0.15*v^2 = 450`

=> `v ~~ 77.45` m/s

If the arrow is shot vertically upwards it rises to a height h where the potential energy of the arrow is equal to the gravitational potential energy. This gives: `0.15*9.8*h = 450`

=> `h ~~ 306.12` m

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