J
Justyn H
Guest
Well... I'm assuming speed is a scalar quantity, not a vector, because negating drag, horizontal velocity should remain constant and vertical velocity is initally zero and so can never mathematically double. So, assuming speed is scalar and not vector (i.e. not dependant on direction):
(a) Speed initial: 60 m/s
Speed doubled: 120 m/s = speed initial + (9.81 m/s^2 * t)
120 = 60 + 9.81t
60 = 9.81t
t = 60 / 9.81
(b) Speed doubled again: 240 m/s
240 = 60 + 9.81t
180 = 9.81t
t = 180/ 9.81
(a) Speed initial: 60 m/s
Speed doubled: 120 m/s = speed initial + (9.81 m/s^2 * t)
120 = 60 + 9.81t
60 = 9.81t
t = 60 / 9.81
(b) Speed doubled again: 240 m/s
240 = 60 + 9.81t
180 = 9.81t
t = 180/ 9.81