Question

A body is projected with velocity u making an angle with teta with horizontal. It's velocity when it is perpendicular to the initial velocity vector is

Answer 1

a projectile moves with speed u and inclined with horizontal is ∅
then, initial velocity vector (U)
= ucos∅i + usin∅ j

let at time = t velocity of projectile is perpendicular upon initial velocity .
then, that time velocity ( V)= ucos∅ i + ( usin∅ -gt) j

now,
U and V are perpendicular so,
dot product of U and V = 0

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

u² cos²∅ + u²sin²∅ -usin∅gt = 0
u²( cos²∅ + sin²∅ ) = usin∅gt
u = sin∅ gt

t = u/gsin∅

now , put this in V
V = ucos∅i + ( usin∅ -gu/gsin∅)j

= u cos∅i + u( sin²∅ -1)/sin∅ j
= ucos∅i - ucos²∅/sin∅ j

magnitude of V

V = u √( cos²∅ + cos²∅cot²∅)
u√( cos²∅(1 + cot²∅))
=u√cos²∅(cosec²∅)
=ucot∅

hence , V = ucot∅

Answer 2

Answer:

here's u'r answer......,