show that the path followed by a projectile is a parabola
Answers
Answer:
Explanation:
♤A body or object upwards at an angle theta other than 90° with horizontal is called projectile.
♤Let projectile is thrown with a initial velocity u at an angle theta there with horizontal.
U=uxi+yuk
Ux=ucos theta
Uy=usin theta
Neglecting air resistance
Along x direction
S=ut+1/2at^2
Distance covered by projectile = velocity time
t=X(x)/ucos theta ----------(1)
Along y direction
S=ut+1/2at^2
y=usin theta × t-1/2gt^2 ---------(2)
Substituting (1) in (2)
y= usin theta × x/ucos theta - 1/2 g (x/ucos theta)^2
y=x tan theta - (g/2u^2cos^2theta
Let tan theta =A
-g/2u^2cos^2 theta =B
Y= Ax+Bx^2--------(Parabola)
This equation represents equation of parabola.
Hence path of projectile(Trajectory) is a parabola....
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