Question

A particle is projected from the ground at an angle of 30° with the horizontal with an initial speed of 20m/sec . At what time will the velocity vector of the projectile be perpendicular to the initial velocity (in seconds)?​

Answer 1

Answer:

A particle is projected from the ground at an angle of 30° with the horizontal with an initial speed of 20m/sec. The velocity vector of the projectile is perpendicular to the initial velocity at the time t = 4s.

Explanation:

Given,

The angle of projection (θ) = $30^{0}$

Initial velocity (u)   = 20m/s

Let velocity vector of the projectile and initial velocity vector are perpendicular,  

from the figure,

α = 90-30 =$60^{0}$

v cos α = u cosθ

v =   $\frac{20\times cos30}{cos60} =\frac{20\times\frac{\sqrt{3} }{2} }{\frac{1}{2} }$ = $20 \sqrt{3} m/s$

By the equation of motion, $v_{x}$= $u_{x}$+at

where acceleration a = -g

$v_{x}=u_{x} - gt$

$u_{x}$ = x component of u = usin30= $20\times \frac{1}{2} = 10$ m/s

$v_{x}$ =  x component of v

    =  vsin30

    =  $20\sqrt{3} \times \frac{\sqrt{3} }{2} = -30 m/s$  (Taking as negative since moving down)

above equation becomes,

$u_{x} - gt= v_{x}$

$10- 10t =-30\\t= \frac{40}{10} =4s$