Simple physics question explain your answer?

Question

It is possible to shoot an arrow at a speed as high as 105 m/s.

(a) If friction is neglected, how high would an arrow launched at this speed rise if shot straight up?

Answer 1

You have to assume that the arrow accelerates at a constant rate due to gravity (-9.8 m/s^2). You also know that at its peak height the velocity of the arrow is 0 because it has to stop before it can change directions.

Use the kinematic equation: (vf)^2 = (vi)^2 + 2*a*d

Where vf=0m/s, vi=105m/s, and a=-9.8m/s^2.

d=((vf)^2 -(vi)^2)/(2a)

=562.5 m

Answer 2

Given: v0= 105 m/s, g= - 9.81 m/s^2, vf=0 m/s, y0= 0 m

v0= initial velocity, g=gravity, vf = final velocity, y0=starting height, yf= final height, ^2=squared

Note that final velocity is equal to zero, because at the max height the arrow is neither going up or falling down.

vf^2-v0^2=(2)(g)(yf-y0)

(0 m/s)^2-(105 m/s)^2=(2)(-9.81 m/s^2)(yf-0 m)

-11025 m^2/s^2=(-19.62 m/s^2)(yf)

(-11025 m^2/s^2) / (-19.62 m/s^2) = yf

562 m = yf

Answer 3

j13562 is right and should get the points :) But just for info, you can find the same problem plus the solution in detail at:

http://209.85.135.104/search?q=cache:4_d...

(In the one in the link, the velocity was 100 m/sec but the math is the same. It also gives you the solution for its time in flight. That site also has answers to some other questions too.)

EDIT: mecdub is right too, by the looks...(Her answer came up while I was posting mine.) The only difference is whether you use 9.81 or 9.80 m/s(/s) for the value of g. I think most accept 9.8 for more straightforward calculations like this one.

Oh, part b)? If I've got my sums right it's just 2v0/g. ie, 210/9.8, which is 21.43 seconds to nearest 1/100th of a sec.