Question

A projectile is given an initial velocity of
(3+2j)m/s. The cartesian equation of its
path considering vertical plane as xy plane, is
(g = 10 m/s)
1) y = 2x - 5x2
(2) y = x - 5x2
(3) 4y = 2x - 5x
(4) 9y = 6x - 5x​

Answer 1

Explanation:

U=(

i

^

+2

j

^

)m/0

Ux=

i

^

m/0

x=S

x

=Ux⋅t

x=S

x

=1t=t …………..(ii)

U

y

=2

j

^

m/0

y=S

y

−Uyt+

2

1

at

2

S

y

=2t−

2

1

×10t

2

y=S

y

=2t−5t

2

………..(ii)

From equation (i)

y=2(x)−5x

2

5x

2

−2x+y=0