 #1
BurpHa
 44
 13
 Homework Statement:

A fire hose held near the ground shoots water at a
speed of 6.5 m / s. At what angle(s) should the nozzle point
in order that the water land 2.5 m away?
 Relevant Equations:
 X = V0 * t + 1/2at^2. t = 2V0y / 9.8
We know the time it takes the water complete the whole parabola is (sin(x) * 6.5 * 2) / 9.8.
So I come up with (sin(x) * 6.5 * 2) / 9.8 * cos(x) * 6.5 = 2.5, because the x component of the velocity is the same for the whole time.
But I get the results like these: x≈0.30929171+πn,1.26150461+πn.
I see the correct results are x=72.27889027+180n,17.72110972+180nx
When I use t = 2.5 / cos(x) * 6.5 and plug it in 0 = sin(x) * 6.5 * (2.5 / cos(x) * 6.5)  4.9 * (2.5 / cos(x) * 6.5) ^ 2, I get the desired results.
I don't know what is not right in (sin(x) * 6.5 * 2) / 9.8 * cos(x) * 6.5 = 2.5 that gets me wrong.
So I come up with (sin(x) * 6.5 * 2) / 9.8 * cos(x) * 6.5 = 2.5, because the x component of the velocity is the same for the whole time.
But I get the results like these: x≈0.30929171+πn,1.26150461+πn.
I see the correct results are x=72.27889027+180n,17.72110972+180nx
When I use t = 2.5 / cos(x) * 6.5 and plug it in 0 = sin(x) * 6.5 * (2.5 / cos(x) * 6.5)  4.9 * (2.5 / cos(x) * 6.5) ^ 2, I get the desired results.
I don't know what is not right in (sin(x) * 6.5 * 2) / 9.8 * cos(x) * 6.5 = 2.5 that gets me wrong.