TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Velocity of the projectile, v=140m∕s
The angle of projection, θ=60∘.
The horizontal component of the velocity is always constant throughout the motion.
Let the velocity is u when the angle is 30∘ with horizontal.
∴‌‌v‌cos‌60∘=u‌cos‌30∘
⇒‌‌‌

v

2

=‌

√3u

2

⇒u=‌

v

√3

=‌

140

√3

m∕s
For vertical motion, uy−vy=gt
⇒‌v‌sin‌60∘−u‌sin‌30∘=gt
⇒‌‌

√3v

2

−‌

140

√3

×‌

1

2

=gt
⇒‌gt=‌

140√3

2

−‌

140

2√3

=‌

140

2

(√3−‌

1

√3

)
⇒‌t=‌

70(3−1)

√3×10

=‌

14

√3

S‌‌(∵g=10m∕s2)