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+vyj v2=u‌cos‌θ‌+(u‌sin‌θ−gt) [∵vy=uy−gt=u‌sin‌θ−gt].....(ii)
As, velocities are perpendicular,
50v1â‹…v2=0 From Eqs (i) and (ii), we get
⇒u2cos2θ+u‌sin‌θ(u‌sin‌θ−gt)=0 ⇒u2cos2θ+u2sin2θ−u‌sin‌θ×gt=0
⇒u2(cos2θ+sin2θ)−ugt‌sin‌θ=0 ⇒‌‌u2=ugt‌sin‌θ‌‌(∵sin2θ+cos2θ=1)
⇒‌‌t=‌ Here,
u=15m/s,g=10m/s2 and
θ=30∘