Question

Derive an expression for velocity of a projectile at any instant of time

Answer 1

Net velocity at any instant of time t

 



 

 

 

 

 

We know ,at any instant t

vx= u
vy= gt
v= (vx2 + vy2)1/2 = [u2 + (gt)2]1/2

 

Direction of v with the horizontal at any instant :

(angle) = tan-1 (vy/vx)= tan-1 (gt/u)

Time of flight (T):
It is the total time for which the projectile is in flight ( from O to B in the diagram above)

To find T we will find the time for vertical fall 
From y= uyt + (1/2) gt2
When , y= h , t=T

h= 0 + (1/2) gt2

T= (2h/g)1/2

Range (R) :
It is the horizontal distance covered during the time of flight T.

From x= ut 
When t=T , x=R

R=uT

R=u(2h/g)1/2

Answer 2

x = u cosθ × t

 

t = x/(u cosθ)                                                   ...... (1)

 

The vertical distance travelled by projectile is given from kinematical equation s = ut + ½at2.

Here, s = y, a = ay = -g, u = uy = u sinθ

 

Thus, we have

 

y = u sinθ - ½gt2                                              ...... (2)

 

Substituting equation (1) in (2), we get

y=xtanθ−gx^2/2v^2cos^2(θ)