TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

The given projectile motion is as shown in figure, Let t be the time after which the initial direction is perpendicular to its direction of motion.
As, velocity of a projectile has two components So, its initial velocity is given by
v1=u‌cos‌θ‌hat‌i+u‌sin‌θ‌hat‌j.....(i)
Since, in projectile motion the horizontal component of velocity remains same, while the vertical component varies, so its velocity after time t‌, ‌
v2=uy

∧

i

+vyj
v2=u‌cos‌θ‌

∧

i

+(u‌sin‌θ−gt)

∧

j

[∵vy=uy−gt=u‌sin‌θ−gt].....(ii)
As, velocities are perpendicular, 50v1⋅v2=0 From Eqs (i) and (ii), we get
⇒u2cos2θ+u‌sin‌θ(u‌sin‌θ−gt)=0
⇒u2(cos2θ+sin2θ)−ugt‌sin‌θ=0
⇒‌‌t=‌

u

g‌sin‌θ

Here, u=15m/s,g=10m/s2 and θ=30∘