TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Given, initial speed of particle, v=(3

∧

i

)‌m∕s
Acceleration of particle, a=(−1

∧

i

−0.5

∧

j

)‌m∕s2
As we know that,
sx‌‌=vxt+‌

1

2

axt2. . . (i)
and sy‌‌=vyt+‌

1

2

ayt2 . . . (ii)
where, vx,vy is the initial speed of particle along X,Y-axes respectively, t be the time taken by the particle to reach maximum height.
∴‌‌sx=3t−‌

1

2

(1)t2
On differentiating both sides w.r.t t, we get
‌

dsx

dt

=3−t=vx′
where, vx′ is final speed of particle along X-axis.
∵ At maximum distance\/ height, speed becomes 0ms−1
⇒‌‌t=3s
∴‌‌sx=3×3−‌

1

2

(1)×32
=9−‌

9

2

=‌

9

2

m
sx=‌

9

2

∧

i

m
and ‌‌sy=uyt−‌

1

2

ayt2
⇒‌‌sy=0×3−‌

1

2

×‌

1

2

×32
=‌

−9

4

m
∴‌‌sy=−‌

9

4

(

∧

j

)
∵ Particle reaches to maximum height and returns.
∴‌‌sy=−‌

9

4

×2

∧

j

=−‌

9

2

∧

j

∴‌‌s=sx

∧

i

+sy

∧

j

=‌

9

2

∧

i

−‌

9

2

∧

j

=‌

9

2

(

∧

i

−

∧

j

)