Show the part of the projectile is parabola in the case of oblique projection
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Suppose the projectile is lunched with an initial velocity of v0 at an angle θ from the horizontal (x). Then the horizontal and vertical component of the velocities are
vx=v0cosθ and vy=v0sinθ
accelerations are
ax=0 and ay=−g
Now the position at any instant t is given by
x=v0tcosθ.....................(1)
y=v0tsinθ−12gt2.....................(2)
Now from equation (1) we can find that t=xv0sinθ, put that value on equation (2) then we can get
y=x tanθ−g2v02cos2θx2....................(4)
Now equation (4) is like equation y=ax+bx2 where a=tanθ and b=g2v02cos2θ. This proves the projectile will be a parabola.
vx=v0cosθ and vy=v0sinθ
accelerations are
ax=0 and ay=−g
Now the position at any instant t is given by
x=v0tcosθ.....................(1)
y=v0tsinθ−12gt2.....................(2)
Now from equation (1) we can find that t=xv0sinθ, put that value on equation (2) then we can get
y=x tanθ−g2v02cos2θx2....................(4)
Now equation (4) is like equation y=ax+bx2 where a=tanθ and b=g2v02cos2θ. This proves the projectile will be a parabola.
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