BITSAT 2023 Shift 2 Solved Paper

Given, Initial velocity of projectile, u=40m∕ s
Angle, θ=30∘
Time of flight
T=‌

2usin‌θ

g

=‌

2×40×1

10×2

=45‌‌(∵g=10m∕ s2)
It means projectile is at maximum height at t= 2 s. At maximum height vertical component of velocity is zero.
Velocity at t=2 s=Vx=u‌cos‌θ=40‌cos‌30∘
=20√3ms−1‌. ‌