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  • A particle is thrown with 10m/s at an angle of  with horizontal the time at which its velo

A particle is thrown with 10m/s at an angle to 60^{\circ} with horizontal. The time at which its velocity is perpendicular to the initial velocity is ( g = 10 m/s )

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Given- 

  • Initial velocity, u = 10 m/s
  • Angle of projection, θ = 60°
  • Gravitational acceleration, g = 10 m/s2

Condition for Perpendicularity - The velocity is perpendicular to the initial velocity when the dot product of the initial velocity vector and the velocity vector at time t is zero (u⋅v=0) uxvx+uyvy=0

  1. In x- axis, ux = vx = ucosθ = 10 x 0.5= 5 m/s
  2. In y- axis, uy = usinθ = 10 x sin 60= 8.66 m/s and vy = uy −gt = 8.66- 10t

uxvx+uyvy=0→5×5+8.66×(8.66−10t)=0 Simplyfing the equation, we get t = 1.15 seconds

Therefore, the time at which the velocity is perpendicular to the initial velocity is approximately 1.15 seconds.

Posted by

Saniya Khatri

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