Question

prove that the path of projectile is parabola in nature

Answer 1

Answer:

Let a body is projected with speed um/s inclined θ 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

2

θ

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

Answer:

l

Explanation:

Let a body is projected with speed um/s inclined θ 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

2

θ

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.