deriving the range equation or, how to get there from here
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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
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