an arrow flies to the moon with an initial velocity of 55m/s. Its height after t seconds is given by H=55t-0.83t^2. WIth what velocity will the arrow hit the moon?
2006-09-24 12:14:52 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(v^2-v0^2)=2as
s=(v^2-v0^2)/(2a)
from the equation above, a=-1.66m/sec^2
s=55*55/2/1.66
s=911 m
The Moon is 238,000 miles away. That arrow ain't gonna come close.
Now, if the arrow leaves the Moon with a vertical velocity of 55 m/s, it will hit the moon with a velocity of -55 m/sec.
(up being +)
2006-09-24 12:26:34 · answer #1 · answered by Helmut 7 · 1⤊ 0⤋
If by height you mean height above the moon, the arrow hits the moon when H = 0, which occurs at time t given by
0 = 55t - .83*t^2; from this t = 55/.83
The acceleration (by inspection) is 2*.83, since s = v*t - .5*a*t^2
The velocity will be 55 + a*t or 55+2*.83*55/.83 = 3*55m/s
This answer bothers me, but I can't see anything wrong with the analysis based on my assumption about H. If I am wrong, hopefully someone will correct me.
2006-09-24 19:27:38 · answer #2 · answered by gp4rts 7 · 1⤊ 0⤋
Since you have the height equation given in terms of the Moons gravity I'm guessing that the arrow was shot upwards from the surface of the Moon.
Here's a simple fact: Neglecting air resistance, an object will hit the ground with *exactly* the same velocity as it was launched upwards. And that holds true on *any* planet
The reason is that, when launched, an object has some amount of kinetic energy. At the 'top' of its path (when its velocity is zero) all of that kinetic energy has become potential energy and, as it falls, all of that potential energy is converted back to kinetic energy.
So your arrow hits the Moon at 55 m/s.
Doug
2006-09-24 19:21:33 · answer #3 · answered by doug_donaghue 7 · 1⤊ 0⤋
This is a trick question. The arrow will never hit the moon.
2006-09-24 20:16:23 · answer #4 · answered by Yahoo! Answerer 6 · 0⤊ 1⤋
Never, because 55M/sec comes nowhere near escape velocity
2006-09-24 19:19:10 · answer #5 · answered by prusa1237 7 · 0⤊ 1⤋
Difficult - glad I don't have to answer the question.
Good Luck though
2006-09-24 19:16:48 · answer #6 · answered by Anonymous · 1⤊ 0⤋
this question cannot be solved (in Math context - not taking into consideration that it's not possible)..enough information is not given..
# I don't think what Doug has said is correct.
2006-09-24 19:25:47 · answer #7 · answered by Rollester 4 · 0⤊ 1⤋
sorry I have no clue, but if you fine out let me know.
2006-09-24 19:35:32 · answer #8 · answered by ? 5 · 1⤊ 0⤋
how high is the moon???????????
2006-09-24 19:16:29 · answer #9 · answered by bobby_yeah 2 · 0⤊ 1⤋