Question

An archer puts a 0.300kg arrow into a bow and draws the bowstring back. She exerts an average force of 201N to draw the bow string back 1.30m
a. Assuming no frictional loss with what speed does the arrow leave the bow?
b. How high will it rise if the arrow is shot straight up?

(I really need help.. I'm doing module work at my school by myself and I was told to go ask my science instructor but she refuses to give me simple help while and guidance while if i ever get stuck on a problem. PS. She's just sitting at her desk eating chips and watching netflix on her computer while the class she has is working. so the help would really be appreciated!)

Answer 1

Well sorry to hear about your potato head Netflix watching "teacher".  But here we go! For part (a) we have:

(a)

m=0.3kg,  F=201 N, $\Delta x=-1.30$ this is negative because our if we think of an undrawn bow as a zero, pulling it back 1.30 meters means -1.30 meters.  First we will solve for the acceleration by using F=ma and so:

$F=ma\\\\a=\frac{F}{m}=\frac{201N}{0.3kg}=670m/s^2$

Next we will use the following kinematic equation to solve for our initial velocity, note that our final velocity is zero since eventually it will stop and so:

$v_{final}^2=v_{initial}^2+2ad$

where a is acceleration, and d is distance our bow was drawn and so:

$0=v_{initial}^2+2(670)(-1.30)\\\\1742=v_{initial}^2\\\\\sqrt{1741}=v_{initial}\\41.74m/s=v_{initial}$

So the arrow leaves the bow at approximately 41.74 m/s (that's about 91.71 miles per hour ouch).

(a) Initial velocity = 41.74 m/s

(b)

Next If the arrow is shot an arrow straight up we have to worry about gravity (g = -9.8 m/s^2) "pushing" the arrow down as your arrow goes up higher and higher (ignoring air resistance) we have:

$v_{final}=v_{initial}+at$

Note that the final velocity in this system is zero, at that point the arrow will come right down since it has no more velocity upwards therefore we have:

$0=41.74+(-9.8)t$

$0-41.74=41.74-41.74+(-9.8)t\\\\-41.74=-9.8t$

$\frac{-41.74}{-9.8}=\frac{-9.8t}{-9.8} \\\\4.26=t$

Therefore it will take 4.26 seconds for the arrow to stop and return back down, meaning that the arrow will reach it's max height at that time.  Finally we can figure out what that height is by using another kinematic equation:

$h=v_{initial}t+\frac{1}{2}gt^2$

Where g=-9.8, initial velocity=41.74 m/s and t=4.26 seconds and so:

$h=(41.74)(4.26)+\frac{1}{2}(-9.8)(4.26)^2\\\\h=88.89m$

So the arrow will reach a max height of 88.89 meters (about 292 ft).

(b) 88.89 meters