Projectile Motion

Class XI

Derivation of Projectile Equation

An object that is in flight after being thrown or project is called a projectile.

Consider about an object that is projected with an angle θ from horizon. The velocity in which the object is projected is v.

Velocity along vertical = vsinθ
Velocity along horizon = vcosθ
Time of flight of the projectile = T
Maximum Height of the projectile = H
Horizontal Range of the projecile = R
Time ascent = Time of decent = t
Time of flight = T = Time of ascent + Time of decent = t + t = 2t

Maximum distance covered by projectile along vertical direction = Y

Maximum distance covered by projectile along horizontal direction =X

Therefore,

Y = vsinθ.T - gT²/2  -------------------- (i)

X = vcosθ.T

T = X/vcosθ

Now putting the value of T

Y = vsinθ.X/vcosθ - gX²/2cos²θ

Υ = X.tanθ - gX²/2cos²θ ------------------------------ (ii)

Τhis is known as equation of Trajectory. This represent the equation of parabola.

Hence the path of the projectile is a parabola.

Maximum height of a projectile (H).

Here time of ascent = t = time of flight/2 = T/2 (Time taken by projectile to attains maximum height)

At maximum height the velocity of the projectile = 0

Now from kinematic equation

0 = vsinθ - gt

t = vsinθ /g

Time of ascent = Time of decent = t = vsinθ /g

Time of flight = T = 2vsinθ /g

Therefore, 

H = vsinθ.t - gt²/2

Now putting the value of t.

H = (vsinθ )² /g - g(vsinθ)²/2g²

H =  (vsinθ )² /2g 

This is the equation of maximum height.

Horizontal Range = R = vcosθ.T

Now putting the value of T

R = vcosθ.2vsinθ /g

R = v²sin2θ/g

Important equations derive from projectile are

1. Υ = X.tanθ - gX²/2cos²θ
2.  t = vsinθ /g
3. T = 2vsinθ /g
4. R = v²sin2θ/g 

Integration By Parts

Chapter 7

Class XII

Exercise 7.6

Q.No. 1. x.sinx

Solution:

From question it is clear that

I = ഽx.sinx.dx

Put First function = x
Second function = sinx
Now, applying integration by parts;
I = - x.cosx + ഽcosx.dx
or, - x.cosx + sinx + C Ans.

Q.No. 2. x.sin3x

Solution:

From question it clear that

I = ഽx.sin3x.dx
Put first function = x
Second function = sin3x
Now applying integration by parts
I = - x.cos3x/3 + ഽcos3x.dx/3
or, -x.cos3x/3 + sin3x/9 + C Ans.

Q.No. 3. x².ex

Solution:

From question it is clear that
 I = ഽx².ex.dx

Put first function = ex
Second function =  x²

Now applying integration by parts
I =  x².ex - ഽ2x.ex.dx
Again applying integration by parts
or,x².ex - 2x.ex +  ഽ2ex.dx
or,x².ex - 2x.ex + 2ex + C
or, ex(x² - 2x + 2) + C Ans.

Q.No. 4. x.logx

Solution:

From question it is clear that
I = ഽx.logx.dx
Put first function = logx
Second function = x

Now, applying integration by parts

I = x².logx/2 - ഽx.dx/2
or, x².logx/2 - x²/4 + C Ans.

Q.No. 5. x.log2x

Solution:

From question it is clear that
I = ഽx.log2x.dx

Put first function = log2x
Second function = x

Now applying integration by parts

I = log2x.x²/2 - ഽx².2/4x.dx
or, log2x.x²/2 - x²/4 + C Ans.

Q.No. 6. x².logx

Solution:

From question it is clear that
I = ഽ x².logx.dx

Put first function = logx
Second function = x²

Now applying integration by parts

I = logx. x³/3 - ഽx²dx/3
or, logx.x³/3 - x³/9 + C Ans.

Q.No. 7. xsin⁻¹x

Solution:

From question it is clear that
I =  ഽxsin⁻¹x.dx

Put first function = sin⁻¹x
Second function = x

Now applying integration by parts

I = sin⁻¹x. x²/2 - ഽx².dx/2√1-x²

or,  sin⁻¹x. x²/2+ഽ(1-x²-1).dx/√1-x²
or,  sin⁻¹x. x²/2+ ഽ√1-x².dx/2 -  ഽdx/2√1-x²
or,  sin⁻¹x. x²/2+ √1-x²/4 + sin⁻¹x./4 - sin⁻¹x./2 + C
or, (2x² - 1)sin⁻¹x./4 + √1-x²/4 +C Ans.

Q.No. 8 x.tan⁻¹x

Solution:

From question it is clear that
I = ഽx.tan⁻¹x

Put first function = tan⁻¹x
Second function = x

Now applying integration by parts

I = x².tan⁻¹x/2 - ഽx².dx/2(1 + x²)
or, x².tan⁻¹x/2 -ഽ(1 + x² - 1).dx/2(1 + x²)
or, x².tan⁻¹x/2 -ഽdx/2 + ഽdx/2(1 + x²)
or, x².tan⁻¹x/2 - x/2 + .tan⁻¹x/2 + C Ans.

Rest questions will be solved latter.