For all of you Mac-Heads out there...NA
- Thread starter vontresc
- Start date
tonycondon
Gastons CRO (Chief Dinner Reservation Officer)
1) yes, he'll have $1,024, so he'll be 424 long.
2) no, by the time he gets the PDF downloaded, he'll be 1.5 feet into the intersection, smack dab in middle of bus.
3) 45.23 meters, neglecting aerodynamic forces.
2) no, by the time he gets the PDF downloaded, he'll be 1.5 feet into the intersection, smack dab in middle of bus.
3) 45.23 meters, neglecting aerodynamic forces.
Last edited:
wbarnhill
Final Approach
Moral of the story: If you don't want to be crushed by a bus, choose a phone that does EVDO.
vontresc
En-Route
Hmm I got 86.45m for the distance...please correct my calculations if I made any errors.
1) at a launch angle of 30 degrees and an initial velocity os 22 m/s that gived us a vertical vector of 11 m/s and horizontal vector of 19 m/s
2) calculate time to travel to the top of the thrown arc d = vt = .5at^2 Therefore t = 2v/a = 22 m/2 /9.81 m/s^2 = 2.24sec
3) absolute height over ground = .5 at^2 + launch height = 24.67m + 1.5m = 26.17 m
4) Time from top of arc to impact = (2d/a) ^.5 = 2.31 sec
5) distance travelled = time up + time down + horizontal speed = 86.45m
Of course this is neglecting any erodynamic forces....
Pete
1) at a launch angle of 30 degrees and an initial velocity os 22 m/s that gived us a vertical vector of 11 m/s and horizontal vector of 19 m/s
2) calculate time to travel to the top of the thrown arc d = vt = .5at^2 Therefore t = 2v/a = 22 m/2 /9.81 m/s^2 = 2.24sec
3) absolute height over ground = .5 at^2 + launch height = 24.67m + 1.5m = 26.17 m
4) Time from top of arc to impact = (2d/a) ^.5 = 2.31 sec
5) distance travelled = time up + time down + horizontal speed = 86.45m
Of course this is neglecting any erodynamic forces....
Pete
tonycondon
Gastons CRO (Chief Dinner Reservation Officer)
i think you found the time to get back to starting elevation right. then you find the time to fall the 1.5 meters from that.
-1.5 = 22sin(-30)*t-.5*9.8*t^2
comes to .129 seconds
so total time of flight is 2.374 seconds
horizontal velocity is 22cos(30) = 19.05 m/s
distance horizontally is about 45 meters
-1.5 = 22sin(-30)*t-.5*9.8*t^2
comes to .129 seconds
so total time of flight is 2.374 seconds
horizontal velocity is 22cos(30) = 19.05 m/s
distance horizontally is about 45 meters
flyingcheesehead
Taxi to Parking
75 KB / 9KB/s = 8 1/3 sec.
8 1/3 * 4.5 ft/sec = 37.5 feet, so he'll be 1.5 feet into the intersection. However, he won't be hit by the bus because either it's moving and it'll already be gone before he gets there 8 1/3 seconds later, or it's not moving so he will just walk right in front of it.
Last edited:
tonycondon
Gastons CRO (Chief Dinner Reservation Officer)
kent, your normally amazing mental math skills have failed you.
75/9 is 8 1/3
75/9 is 8 1/3
flyingcheesehead
Taxi to Parking
Gaah. That's what I get for doing "reverse math." 75=9*(9-2/3)