Question

explain step by step please

1) A projectile returns to its original height after 4.08 seconds, during which time it travels 76.2 meters horizontally. If air resistance can be neglected, what was the projectile's initial speed?

(Use g = 9.80 m/s2)

Answer 1

Given

t = 4.08 s
R = 76.2 m
g = 9.80 m/s^2

First I found the vertical component of the initial velocity through using the equation meant to solve for the time to peak height. 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 that it 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 that time, because that number had no bearing on the problem. I moved on, solving for the initial horizontal component of velocity, Vox using 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 elevation of launch and impact are the same, and no air resistance, then Vox and Voy will be the same. So then we can solve for the velocity at impact:

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