AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper

Given info:
The initial velocity of a particle is (

^

i

+2

^

j

)ms−1.
Here, the initial velocity of a particle in x direction is ux=1ms−1 and the initial velocity of a particle in y directionis uy=2ms−1.
So, the acceleration of a particle in x direction is ax=0 and the acceleration of a particle in y direction is ay=−10ms−2.
The horizontal distance covered by projectile in time t is,
x=uxt
=(1ms−1)t
=t
The vertical distance covered by projectile in time t is,
y=uyt+

1

2

ayt2
=(2ms−1)t+

1

2

(−10ms−2)t2
=2t−5t2
y=2x−5x2