Question

A body is projected horizontally with a speed v.Find the velocity of the body when it covers equal distances in horizontal and vertical directions.

Answer 1

Answer:v=√(5v)

Explanation:Initial velocity =V

We know the body is projected horizontally so velocity component in vertical direction will be zero.

Now for calculating horizontal distance and vertical distance we have to calculate the height and the distance

So,

h(y)=vt + 1/2at2

Since it's falling down initial velocity (v) will be zero and because it's falling instead of a we'll write a=-g

Therefore,

h(y)= -gt2 ____________(1)

Now for calculating horizontal distance we have to find the distance.

So,

S(X)=vt____________(2)

Acc. to question, we have to find the velocity when the body covers equal distance in horizontal and vertical ,so we will equate (1) & (2)

h(y)=S(X)

gt2/2= vt

We'll get

t=2v/g

Now applying the equation

v(y)= u(y)-gt

v=-g2v/g

v(y)=2v

Now in vector form

|V| = √(Vx)2 + (Vy)2

=√V2 + 2V2

=√V2+ 4V2

=√5V2

=√5 V

Answer 2

Answer:

$V_r=v\sqrt 5$

Explanation:

Given that

Horizontal speed = v m/s

The horizontal distance cover in time t

R= v .t

The vertical distance cover in time t

$h=\dfrac{1}{2}g\ t^2$

given that both distances are equal

R = h

$v.t=\dfrac{1}{2}g\ t^2$

2 v = g t

t= 2 v/g

The vertical velocity after time t

V= g t= g x  2 v/g

V= 2 v

The horizontal speed= v

The resultant speed Vr

$V_r=\sqrt{(2v)^2+v^2}$

$V_r=v\sqrt 5$