The terminal velocity is vT When = 0 the object reaches its maximum height (up is negative) v xup The time tup to the maximum height The calculations agree with the values for and determined from the tup xup graphs. From the graphs: = 0 time = 2.275 s velocity = 11.21 m.s-1 x t v Time to fall from max height to origin = (2.375 1.169) s = 1.206 s tdown – takes slight longer to fall then rise to and from origin to max height Launch = 12.00 m.s-1 slightly greater than return speed = 11.21 m.s-1 In the absence of any resistive forces At maximum height Numerical Approach We can also find the velocity and displacement of the object by solving Newton’s Second Law of motion using a finite difference method. We start with (4) In the finite difference method we calculate the velocity and displacement at v x discrete times tk at fixed time intervals N t
The terminal velocity vT is When v = 0 the object reaches its maximum height xup (up is negative) The time tup to the maximum height The calculations agree with the values for tup and xup determined from the graphs. From the graphs: x = 0 time t = 2.275 s velocity v = 11.21 m.s-1 Time to fall from max height to origin tdown = (2.375 – 1.169) s = 1.206 s takes slight longer to fall then rise to and from origin to max height Launch = 12.00 m.s-1 slightly greater than return speed = 11.21 m.s-1 In the absence of any resistive forces At maximum height Numerical Approach We can also find the velocity and displacement of the object by solving Newton’s Second Law of motion using a finite difference method. We start with (4) In the finite difference method we calculate the velocity v and displacement x at N discrete times tk at fixed time intervals t