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