An archer puts a 0.4 kg arrow to the bowstring. an average force of 190.4 n is exerted to draw the string back 1.47 m. the acceleration of gravity is 9.8 m/s². assuming no frictional loss, with what speed does the arrow leave the bow? answer in units of m/s. if the arrow is shot straight up, how high does it rise? answer in units of m.

v = 37.4 m/s , h = 71.39m

Explanation:

To find the velocity given:

m = 0.4 kg

F =190.4 N

d = 1.47 m

g = 9.8 m/s^2

So use the equation of work to solve the kinetic energy

W = F *d = 190.4 N * 1.47m

W = 279.88 J

Ke = 1 / 2 * m* v^2

v = √2*Ke / m =√ 2 *279.88 / 0.4 kg

v = 37.4 m/s

Now to find the high to rise can use the conserved law so:

Ke = Pe

279.88 = m*g*h

Solve to h'

h = 279.88 / 0.4kg * 9.8m/s^2

h =71.39 m

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