Question

the trajectory of a projectile launched from ground is given by the equation y = -0.025x^2 + 0/5 x, wherein x and y are the coordinate of the projectile on a rectangular system of axes. find the initial and angle at which the projectile is launched?

Answer 1

your equation is

\( y = -0.025x^2 + \dfrac{0}{5 x}\)

or maybe

\( y = -0.025x^2 + \dfrac{0}{5 }x\)

and it might be that your answer lies in the formulation \(\tan \theta = \dfrac{rise}{run} = \left. y'(x) \right |_{x = 0} = \left. \dfrac{\Delta y}{\Delta x} \right |_{x = 0}\)

:)

Answer 2

Oh my mistake, that should be 0.5x

Answer 3

I got theta but how can i get the initial velocity.

Answer 4

your approach is currently dimensionless

we can do all sorts of stuff like this: \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{\dot y}{\dot x}\)

velocity requires you to introduce time as a dimension

:(

Answer 5

thank you :)

Answer 6

\[y=0.5x \left( 1-\frac{x}{1/0.05} \right)\]compare it with the general equation for trajectory \[y= x \tan \theta \left( 1- \frac{x}{R} \right)\]