Question

A ball is thrown with velocity 8m/s making an angle 60°with the horizontal.
Its velocity will be perpendicular to the directon of initial velocity of projection after a time of

Answer 1

let velocity of body after time t is perpendicular upon initial velocity .

velocity after time t
v = ucos∅ i + (usin∅ -gt ) j

initial velocity
v° = ucos∅ i + usin∅ j

because v and v° is perpendicular so,

v.v° = 0

{ucos∅ i + (usin∅ -gt) j }{ucos∅ i + usin∅ j } =0

u²cos²∅ +u²sin²∅ -usin∅.gt =0

u - sin∅gt = 0

t = u/sin∅g

=8/sin60.10
=8/√3/210

=1.6√3 sec

t = 1.6√3 sec

Answer 2

First of all. The answer given by Abhi is wrong and needs to be removed.

The real answer+p. -}>

u=8m/s

v=ucos§ +u sin§ -gt

Now u.v=0. {angle is 90°}

So

(ucos§ +u sin§ -gt).(ucos§ +u =0

u-sin$. gt = 0

T= v/sin§g

T= 1.6/root3

Or refer to this image. I hope I helped. Plz mark as brainliest.