3 minutes of coasting/deceleration? That seems like too long.
To gain an additional 13,150 feet at 350mph would take about 25 seconds. 350mph = 1,848,000 feet/hr = 30,800 ft/min = 513 ft/sec.
13,150 feet / 513fps = 25.63 seconds.
I didn't take into account deceleration, because I don't know how, but I don't think the time will change from 25 seconds to 180 seconds. Maybe I'm wrong, but that long of deceleration/coasting time feels off to me.
Yep -- my math was off, too.
Acceleration due to gravity is 32 feet-per-second-per second. Deceleration is nothing but negative acceleration. Acceleration due to gravity is constant (albeit slightly affected by drag). If we ignore drag, we can look at it as how long it would take to go from 0 MPH to 350 MPH. The formula for velocity is V = at. Rearranging that, we get the formula for time: t = V/a
350 MPH is, as you calculated, 513 feet-per-second. Time required to go from 0 to 513 fps (or the opposite) at an acceleration of 32 feet-per-second-per-second is 513 = 32t.
t = 513/32 = 16.03 seconds.
Assuming constant acceleration (which gravity is), the formula for distance is d = 1/2(at^2) We now know the rate of acceleration and the time, so distance is:
d = 1/2(32 x 16.03^2) = 1/2(32 x 256.96) = 16 x 256.96 = 4,111 feet.
BUT ... the article said he reached a height/altitude of 1,875 feet, but he reached a velocity of 350 MPH. Granted, he didn't go straight vertical, but he landed only 1,500 feet from the launch pad. That's less than a third of a mile, so he went pretty much straight up. So either he didn't reach 350 MPH, or else he went a lot higher than 1,875 feet.