Question

96. A body is projected with velocity u making an
angle theta with the horizontal. Its velocity when it
is perpendicular to the initial velocity vector is
a) u sin theta
b) u cot theta
c) u tan theta
d) u cos theta​

Answer 1

Explanation:

ANSWER

Consider theta as alpha

At time t=0 (when body is projected),

Horizontal velocity, u

x

=ucosα

Vertical velocity, u

y

=usinα

Velocity vector,

u

=ucosα

i

^

+usinα

j

^

At any time t;

Horizontal velocity, v

x

=ucosα

Vertical velocity, v

y

=usinα−gt

Velocity vector,

v

=ucosα

i

^

+(usinα−gt

)

^

j

When two vectors are perpendicular, their dot product is zero.

u

.

v

=0

This implies (ucosα)

2

+(usinα)

2

−(usinα×gt)=0

t=

gsinα

u

∴v

y

=usinα−g(

gsinα

u

)=u(sinα−cosecα)

Velocity magnitude, ∣v∣=

((ucosα)

2

+(u(sinα−cosecα))

2

=ucotα