Consider a body projected from a point O on the surface of the Earth with an initial velocity u at an angle θ with the horizontal as shown in Fig.. The velocity u can be resolved into two components

*Motion of a projectile projected at an angle with the horizontal (oblique projection)*

Consider a body projected from a point O on the surface of the Earth with an initial velocity *u* at an angle θ with the horizontal as shown in Fig.. The velocity *u* can be resolved into two components

*ux = u cos **θ** *, along the horizontal direction OX and

*uy = u sin **θ*, along the vertical direction OY

The horizontal velocity *ux* of the object shall remain constant as no acceleration is acting in the horizontal direction. But the vertical component *uy* of the object continuously decreases due to the effect of the gravity and it becomes zero when the body is at the highest point of its path. After this, the vertical component *uy* is directed downwards and increases with time till the body strikes the ground at*B*.

*Path of the projectile*

Let *t1* be the time taken by the projectile to reach the point C from the instant of projection.

Horizontal distance travelled by the projectile in time *t1 *is,

*x *= horizontal velocity x time

*x *= *u cos θ* x* t1 *(or) t1 =* x / u *cos* *θ ... (1)

Let the vertical distance travelled by the projectile in time t1 = *s = y*

At O, initial vertical velocity *u1= u* sin θ From the equation of motion

*s = u1* *t1* *-* 1/2 *gt*12

Substituting the known values,

*y = (u sin θ) t1 -* 1/2 *gt*12 ..(2)

Substituting equation (1) in equation (2),

The above equation is of the form *y = Ax + Bx2* and represents a parabola. Thus the path of a projectile is a parabola.

*Resultant velocity of the projectile at any instant t1*

At C, the velocity along the horizontal direction is *ux* *= u* cos θ and the velocity along the vertical direction is *uy*= *u2.*

*From the equation of motion,*

*u2 = u1 - gt1*

*u2 = u sin θ - gt1*

*∴** The resultant velocity at*

*C is v=root[ux2+u22]*

*v=root[(ucos** θ**)2 + (u sin** θ-gt1**)2]*

*=root[u2+g2t21-2ut1gsin** θ**]*

*The direction of v is given by*

*tan* *α =u2/ux = [ u sin** θ- gt1 ] / [ u cos θ]*

where α is the angle made by v with the horizontal line.

*Maximum height reached by the projectile*

*The maximum vertical displacement produced by the projectile is known as the maximum height reached by the projectile. *In Fig,EA is the maximum height attained by the projectile. It is represented as *hmax.*

At O, the initial vertical velocity (*u1*) = *u* sin θ

At *A*, the final vertical velocity (*u3*) = 0

The vertical distance travelled by the object = *sy* *= hmax*

From equation of motion, *u32* *= u21* *- 2gsy*

Substituting the known values, *(0)* *2= (u sinθ)* *2* *- 2ghmax*

*2ghmax = (u sinθ)* *2*

*hmax = (u sinθ)* *2 / 2g*

*Time taken to attain maximum height*

Let *t* be the time taken by the projectile to attain its maximum height.

From equation of motion *u3 = u1 - g t*

Substituting the known values *0 = u *sin* *θ* - g t*

*g t = u sin *θ

*t** *=* (u *sin* *θ) / g

*Time of flight*

Let *tf* be the time of flight (i.e) *the time taken by the projectile to* *reach B from O through *A. When the body returns to the ground, the* *net vertical displacement made by the projectile

(i.e) *the time of flight is twice the time taken to attain the maximum* *height.*

*Horizontal range*

The horizontal distance OB is called the range of the projectile. Horizontal range = horizontal velocity x time of flight

*(i.e)* *R *=* u *cos* *θ x* tf*

*R = u 2 (2 sinθ cosθ) / g*

*R = (u 22 sin2θ)/g*

*Maximum Range*

From (8), it is seen that for the given velocity of projection, the horizontal range depends on the angle of projection only. The range is maximum only if the value of sin 2θ is maximum.

For maximum range *Rmax *sin 2*θ* = 1

(i.e) *θ* = 45

Therefore *the range is maximum when* *is 45*^{o}*.*

*Therefore the range is maximum when the angle of projection is 45*^{o}*.*

*Rmax =(* *u2 x 1) /g*

*Rmax = u2 /g*

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