PROJECTILE :
Any object or body thrown with an Initial Horizontal Velocity at an angle less than 900 , under the action of gravity,is said to be a Projectile.
* O be the point on the earth’s surface
U be the initial velocity
T be the angle with which the body is projected
* U can be resolved into two components,
(i) UX = U cosT, along the direction OX (Horizontal Velocity)
(ii) UY = U sinT, along the direction OY (Vertical Velocity)
* UX is constant as there is no acceleration acting in the horizontal direction
* UY varies at different points due to gravity. UY decreases as the body rises against gravity,UY is zero at maximum point and UY increases as the body falls till it reaches the ground.
PATH OF THE PROJECTILE :
* t1 be the time taken by the projectile from O to C.
* If x be the horizontal distance then,
x = horizontal velocity (Ux) × time [Ux = U cosT]
i.e. x = U cosT × t1 or t1 = x / U cosT
* If y be the vertical distance and U1=U sinT be the initial vertical velocity at O then
from the equation of motion,
s = ut + 1/2 at2 [s=y; u=U1(i.e.U sinT); t=t1; a=-g]
i.e. y = U1t1 - 1/2 gt12
y = (U sinT)(x/U cosT) - 1/2 g (x/U cosT)2
[sinT/cosT=tanT]
y = x tanT - gx2/2U2cos2T
* The above equation represents the equation of a parabola
(y = Ax+Bx2).so the path of a projectile is a parabola.
RESULTANT VELOCITY OF THE PROJECTILE AT ANY INSTANT t1 :
* At C, if Ux = U cosT is the horizontal velocity and Uy = U2 is the vertical velocity then
from the equation of motion,
v = u + at [v=U2 ; u=U1 ; a=-g ; t=t1]
U2 = U1 - gt1
U2 = U sinT - gt1
* The resultant velocity at C is V2 = Ux2 + U22
V = ( Ux2 + U22 )1/2
V = ( (U cosT)2 + (U sinT - gt1)2 )1/2
V = ( U2 + g2t12 -2Ut1gsinT )1/2
* The direction of V is, tanA = U2/Ux
tanA = (U sinT - gt1) / U cosT
A = tan-1[ (U sinT - gt1) / U cosT ]
A be the angle made by V with Ux.
MAXIMUM HEIGHT REACHED BY THE PROJECTILE (hmax):
The maximum vertical displacement produced by the projectile is known as maximum height reached by the projectile.
* hmax is the maximum height reached by the projectile
* At O, initial vertical velocity U1 = U sinT
* At A, final vertical velocity U3 = 0
From equation of motion, v2 = u2 + 2as
[v=U3 ; u=U1 ; a=-g ; s=hmax]
U32 = U1 - 2ghmax
0 = (U sinT)2 - 2ghmax
hmax = U2 sin2T / 2g
TIME TAKEN TO ATTAIN MAXIMUM HEIGHT :
* tm be the time taken by the projectile to attain maximum height.
From equation of motion,
v= u + at
[ v=U3 ; u=U1 ; a=-g ; t=tm ]
U3 = U1 - gtm
0 = U sinT - gtm
tm = U sinT / g...............(1)
TIME OF FLIGHT (tf):
It is the time taken by the projectile to reach B from O through A.
* tf be the time of flight.
From the equation of motion,
s = ut + 1/2 at2
[ s=sy ; u=U1 ; t=tf ; a=-g ]
* sy = hmax - hmax = 0 ( as the body returns to the ground,net vertical displacement is zero )
sy = U1tf - 1/2gtf2
0 = (U sinT) tf - 1/2gtf2
tf = 2 U sinT / g..............(2)
From equations 1 and 2 , tf = 2tm
* Time of flight is twice the time taken to attain the maximum height.
HORIZONTAL RANGE (R):
The horizontal distance OB is called range of the projectile (R).
Horizontal range = horizontal velocity × time of flight
R = U cosT × tf
R = (U cosT) (2U sinT) / g
R = U2 (2 sinT cosT) / g
R = U2 sin2T / g
MAXIMUM RANGE ( Rmax ):
* The horizontal range depends on the angle of projection. The range is maximum only f the value of sin2T is maximum.
For maximum range Rmax , sin2T = 1
T = 450
* Therefore the range is maximum when the angle of projection is 450.
Rmax = (U2 × 1) / g
Rmax = U2 / g
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