Well sorry to hear about your potato head Netflix watching "teacher". Â But here we go! For part (a) we have:
(a)
m=0.3kg, Â F=201 N, [tex]\Delta x=-1.30[/tex] 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. Â First we will solve for the acceleration by using F=ma and so:
[tex]F=ma\\\\a=\frac{F}{m}=\frac{201N}{0.3kg}=670m/s^2[/tex]
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:
[tex]v_{final}^2=v_{initial}^2+2ad[/tex]
where a is acceleration, and d is distance our bow was drawn and so:
[tex]0=v_{initial}^2+2(670)(-1.30)\\\\1742=v_{initial}^2\\\\\sqrt{1741}=v_{initial}\\41.74m/s=v_{initial}[/tex]
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:
[tex]v_{final}=v_{initial}+at\\\\[/tex]
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:
[tex]0=41.74+(-9.8)t[/tex]
[tex]0-41.74=41.74-41.74+(-9.8)t\\\\-41.74=-9.8t\\\\[/tex]
[tex]\frac{-41.74}{-9.8}=\frac{-9.8t}{-9.8} Â \\\\4.26=t[/tex]
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. Â Finally we can figure out what that height is by using another kinematic equation:
[tex]h=v_{initial}t+\frac{1}{2}gt^2[/tex]
Where g=-9.8, initial velocity=41.74 m/s and t=4.26 seconds and so:
[tex]h=(41.74)(4.26)+\frac{1}{2}(-9.8)(4.26)^2\\\\h=88.89m[/tex]
So the arrow will reach a max height of 88.89 meters (about 292 ft).
(b) 88.89 meters
