Number 6 A projectile returns to it\'s original height after 4.08s,during which
ID: 1678509 • Letter: N
Question
Number 6 A projectile returns to it's original height after 4.08s,during which time it travels 76.2m horizontally. If air resistancecan be neglected, what was the object's initial speed? a. 37.5 m/s b. 27.4 m/s c. 20.4 m/s d. 24.7 m/s e. 37.5 m/s Number 9 A very thin rod, 2.40m long and a weight of 135N, has africtionless hinge at its lower end. It starts out vertically fromrest and fall, pivoting about the hinge. Just as it has rotatedthrough an angle of 55 degrees, the acceleration of the endfarthest from the hinge is closest to: a. 8.03 m/s^2 b. 12.0 m/s^2 c. 48.2 m/s^2 d. 9.80 m/s^2 e. 5.02 m/s^2 Number 6 A projectile returns to it's original height after 4.08s,during which time it travels 76.2m horizontally. If air resistancecan be neglected, what was the object's initial speed? a. 37.5 m/s b. 27.4 m/s c. 20.4 m/s d. 24.7 m/s e. 37.5 m/s Number 9 A very thin rod, 2.40m long and a weight of 135N, has africtionless hinge at its lower end. It starts out vertically fromrest and fall, pivoting about the hinge. Just as it has rotatedthrough an angle of 55 degrees, the acceleration of the endfarthest from the hinge is closest to: a. 8.03 m/s^2 b. 12.0 m/s^2 c. 48.2 m/s^2 d. 9.80 m/s^2 e. 5.02 m/s^2Explanation / Answer
Given t = 4.08 s R = 76.2 m g = 9.80 m/s^2 First I found the vertical component of the initial velocitythrough using the equation meant to solve for the time to peakheight. Since we know time to peak height (half of the given time),I could solve for Voy: tp = Voy / g 2.04 s = Voy / 9.80 m/s^2 Voy = 19.992 m/s Then I used that number to solve for the total maximum height thatit actually achieved (H): H = Voy^2 / 2g H = (19.992 m/s)^2 / (2 * 9.80 m/s^2) H = (399.68 m^2/s^2) / (19.6 m/s^2) H = 20.39 m Then I was disappointed, because it turned out that I'd wasted thattime, because that number had no bearing on the problem. I movedon, solving for the initial horizontal component of velocity, Voxusing the range and new Voy: R = (2*Vox*Voy) / 9.80 m/s^2 76.2 m = [ 2 * (Vox) * (19.992 m/s) ] / 9.80 m/s^2 746.76 = 39.844 * Vox Vox = 18.742 m/s We also know that in the case of a trajectory where the elevationof launch and impact are the same, and no air resistance, then Voxand Voy will be the same. So then we can solve for the velocity atimpact: Vf = SQRT [ (Vfx)^2 + (Vfy)^2 ] Vf = SQRT [ (18.742 m/s)^2 + (19.992 m/s)^2 ] Vf = SQRT [ (351.26 m^2/s^2) + (399.68 m^2/s^2) ] Vf = SQRT [ 750.94 m^2/s^2 ] Vf = 27.4 m/s b is correct
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.