A fire hose held near the ground shoots water at a speed of 6.3m/s. At what angle(s) should the nozzle point in order that the water land 2.5m away? I need help setting up.
Use the following formulas: Δx vi cos(θ) t, where Δx is equal to final - initial in the x position (2.5m - 0m), vi is initial speed, t is time in secs. and Δy vi sin(θ) t + .5 g t^2, where Δy is going to equal zero (it starts and ends on the ground), g is the speed of gravity (-9.81m/s^2). Now just plug in what we know: This is x direction 2.5m 6.5m/s cos(θ) t This is y direction 0.0m 6.5m/s sin(θ) t - 4.905m/s^2 (t^2) so 6.5m/s sin(θ) 4.905m/s^2 (t)t's canceled out get t by itself t1.325s sin(θ) plug t into x eq. 2.5m 6.5m/s cos(θ) 1.325s sin(θ) simplify (units cancel) 0.290 cos(θ)sin(θ) this part is tricky if you forgot trigonometry but notice cos(θ)sin(θ) is equal to [2cos(θ)sin(θ)]/2 and 2cos(θ)sin(θ) is the trig identity sin(2θ)! So now we have [sin(2θ)]/2 0.290 - 2θ 35.484 deg - θ 17.742 deg which is approximately 18 degrees. To find the points on a graph go back to the equations in the beginning and plug in θ 17.742 to find t 0.404 sec. You can find x,y now in whatever increment you'd like. You can also find t by recognizing that t vi cos(θ) / x, gives you same answer. To find the high point, realize that the angle will be 0 at max height. Pretty easy from there. So there's how to find 3 points, you can find more points by playing with different values of t, within 0t.404. I guess you could also plug in 72 degrees the same way to prove that, that is not the correct angle. 18 is correct. Sorry if this is a complicated approach. You did not state what math background you had. Hope it helped. Good luck.
Is this from both sides of the jeep? If you can lock them with the passenger side switch then it is not a relay but the switch itself. Most relay's in the jeep are interchangeable, at least some are on my Grand Cherokee.
