Deriving the Range Equation Or, how to get there from here
Jan 16, 2016
Deriving the Range Equation
Deriving the Range EquationOr, how to get there from here
Keep in mind . . .Horizontal velocity REMAINS CONSTANTNo net force is acting horizontally so there is no horizontal accelerationVertical velocity CHANGESAcceleration due to gravity, ~9.81 m/s2Caused by the unbalanced force of gravity acting on the object
ymaxxxR = 2x
ymaxxxR = 2xqvi
qviviy = vi sin qvx = vi cos q
ymaxxxR = 2xqviviy = vi sin qvx =vi cos q
ymaxxxR = 2xqvi
ymaxxxR = 2xqviAt the top of the path, vfy = 0
ymaxxxR = 2xqviSubstituting in
ymaxxxR = 2xqviSubstituting in
qviviy =vi sin qvx = vi cos qRemember that the initial velocity in the y-direction = vi sin q
ymaxxxR = 2xqvi
The whole point here is to solve for x . . .
Remember that the range, R, = 2x