an archer puts 0.300 kg arrow into a bow and draws the bowstring back. she exerrts and average force of 201N to draw the bow string back 1.30m

a:assuming no friction loss with what speed does the arrow leave the bow?

b:how high will it rise if the arrow is shot straight up?

ughhh :(

Question

an archer puts 0.300 kg arrow into a bow and draws the bowstring back. she exerrts and average force of 201N to draw the bow string back 1.30m

a:assuming no friction loss with what speed does the arrow leave the bow? 

b:how high will it rise if the arrow is shot straight up?

ughhh :(

ujjwal · Answer

Calculate the work done by the archer to draw the bowstring back...
this work done will be equal to initial Kinetic Energy of the arrow given by relation E=1/2 mv^2 ..

So, we have 
Work done by archer = (1/2)mv^2
Calculate v.. That is your answer for first part..

After this i guess you can do the second part yourself..

anonymous · Answer

This is the second time I've typed this as the page collapsed.  Use F = ma to get accelleration.  "201/0.3 =670 ms^-2.  Use v^2 = u^2 +2aS (u=0) so v^2 =2aS = 2 x 670 x 1.3m = 1742.  Take square root  gives 41.7 ms^-1.  For the height we need to find the time (t) in flight; t= v/g (g = acceleration due to gravity = 9.81ms^-2)  41.7/9.81 = 4.25 s. Displacement = 0.5(v) x t  = 0.5 x 41.7 x 4.25 = 88.7 m.