Okay here's a question on my homework sheet and i don't even know where to start on it:
A daring cowboy/girl sitting on a tree limb whishes to drop vertically onto a horse galloping under the tree. The speed of the horse is 10m/s and the distance from the limb to the saddle is 3.0m. What must be the horizontal distance between the saddle and the limb when the cowboy/girl makes his/her move?
As I said i don't even know where to start
2007-11-11 04:50:45 · 5 answers · asked by stefany 1 in Science & Mathematics ➔ Physics
Oh and gravity in this question is 9.8m/s^2
2007-11-11 04:58:27 · update #1
show your work lol
2007-11-11 04:59:50 · update #2
First you need to know how much time the drop takes if the cowboy/girl is to land on the saddle. the saddle would be 3 meters below him/her and as the cowboy/girl drops, his/her inital speed would be zero but there'd be acceleration due to gravity, 9.8 m/s2. Then we have:
d = V0t + (1/2)at^2 but V0=0 so...
d=(1/2)at^2
=>t = sqrt(2d/a)
t=sqrt(2*3/9.8)
t=0.782460796 sec
So it takes the cowperson this long to fall. Now we want the horse to be in the same place is the person at this time. The moment the cow person lets go of the tree, the horse has 0.782460796 seconds to get there, and the horse is going 10 m/s so lets see how far away the horse should be as he/she lets go:
distance = speed*time
distance = 10*0.782460796
distance = 7.82460796 m
So the horse should be 7.8 meters away.
2007-11-11 05:01:45 · answer #1 · answered by Anonymous · 1⤊ 0⤋
well generally fall speed is also 10 m/s iirc...so in order to fall 3 metres before the horse passes the limb the distance needs to be 3 metres....that may sound a bit dodgy but hope it helps :P
2007-11-11 12:56:28 · answer #2 · answered by Anonymous · 0⤊ 1⤋
Please don't try that at home!
The time it takes for that cowboy/girl to fall is used to compute the distance D in question. So since the height h is given...
t=sqrt(2h/g)
D= Vh t
D= Vh sqrt(2h/g)
D= 10 sqrt(2 x 3.0/9.81)=
D= 7.82 m before the saddle.
2007-11-11 12:55:51 · answer #3 · answered by Edward 7 · 0⤊ 1⤋
ok first work out how long it will take the girl to fall 3m
v^2=u^2 +2as
V^2=0^2 +2x9.8x3=58.8
v=7.67m/s
then
v=ut+1/2at^2
7.67=0xt+1/2x9.8xt^2
7.67=4.9xt^2
t^2=7.67/4.9=1.565
t=1.25s
now how far will the horse tavel in 1.25 s =12.5m
so he/she must drop when the horse is 12.5m away
2007-11-11 13:05:16 · answer #4 · answered by Ross C 1 · 0⤊ 2⤋
2.5 METERS
2007-11-11 12:59:02 · answer #5 · answered by Noah 2 · 0⤊ 1⤋