Question

7. Show that the trajectory of an object thrown at certain angle with the horizontal is a parabola,
v2 cinza.
V2 in 20​

Answer 1

Answer:

Two dimensional motion :

→ In two dimensional motion

r

=x

i

^

+y

j

^

→Velocity

v

=

dt

dx

i

^

+

dt

dy

j

^

⇒

V

=V

x

i

^

+V

y

j

^

→O can be resolved into components.

→U

x

=usinθ

U

y

=usinθ

→ After some time

′

t

′

→V

x

=U

x

+a

x

⋅t=ucosθ+cost

⇒V

x

=ucosθ

→V

y

=U

y

+a

y

t

=ucosθ−gt

→

V

=V

x

i

^

+V

y

j

^

→ After some from, displacement

S=ut+

2

1

at

2

X=ucosθt as a

x

=0

Y=usinθt−

2

1

gt

2

→ As X=ucosθ⋅t

t=

ucosθ

x

→ As Y=usinθt−

2

1

gt

2

=usinθ

ucosθ

x

−

2

1

g

u

2

cos

2

θ

x

2

=tanθ×−

2u

2

cos

2

θ

gx

2

⇒Y=Ax−Bx

2

→ where A=tanθ and

B=

2u

2

cos

2

θ

g

and

both are constants

⇒ The above equation represents a parabola.

→ So path of a projectible is parabola.

→ Velocity along X−direction ucosθ always remains constant as there is no gravity in that direction.