1 Answer | Add Yours
Yes we can.
Suppose an objected is projected from ground level in a velocity `V` and at angle `alpha` with the horizontal.
Therefore at t=0,
Velocity in x-direction `= V_x = Vcos(alpha)`
Velocity in y-direction `= V_y = Vsin(alpha)`
`S =ut+1/2at^2` in x-direction.
There is no force acting on x-direction if we neglect air friction.
`x = Vcos(alpha)t+0`
`t = x/(Vcos(alpha))`
`S =ut+1/2at^2` in y-direction.
`y = Vsin(alpha)t - 1/2gt^2`
Substituting for t,
`y = Vsin(alpha) xx x/(Vcos(alpha)) - 1/2g xx (x/(Vcos(alpha)))^2`
`y = xtan(alpha)-(gx^2)/(2V^2cos^2(alpha))`
This can be rearranged as,
`y = -(gx^2sec^2(alpha))/(2V^2)+xtan(alpha)`
sec^2(alpha) = 1+tan^2(alpha)
`y = -(g(1+tan^2(alpha)))/(2V^2) x^2+tan(alpha)x`
Now this is a quadratic expression of x.
We’ve answered 319,819 questions. We can answer yours, too.Ask a question