Question

show that path of projectile is parabola​

Answer 1

Explanation:

with horizontal line.

Then, vertical component of u, u y =ucosθ

Horizontal component of u x =usinθ

acceleration on horizontal, ax=0

acceleration on vertical, ay=−g

Now, use formula

X=u x t

X=ucosθ.t

t=X/ucosθ--------------------------(1)

Again, y=u y

t+1/2a y t 2

y=usinθt−1/2gt 2

put equation (1) here,

y=usinθ×x/ucosθ−1/2gt

2 ×x 2 /u 2 cos 2 θ

=tanθx−1/2gx 2 /u 2 cos 2

y=tanθ.x−1/2gx

2 /u 2 cos 2θ

this equation is similar to standard equation of parabola y=ax

2

+bx+c her, a,band c are constant

So, A projectile motion is parabolic motion.

Answer 2

mujhe nahin Pata I am sorry