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.

Similar Triangles

Exercise 6.4
Class X
NCERT Textbook

Q.No. 1. Let △ ABC ∼△DEF and there areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.

Solution:
ar(ABC)/ar(DEF) = (BC/EF)²
Now, putting the value of areas and EF
64/121 = (BC/15.4)²
Therefore,
BC = 8x15.4/11
Hence,
BC = 11.2 cm Ans.

Q.No. 2. Diagonals of trapezium ABCD with AB॥DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Solution:

Consider in triangles AOB and COD.
<OAB = <OCD     (Alternate angles)
<OBA = <ODC      (Alternate angles)
<AOB = <COD      (Vertically opposite angles)
△AOB∼ △COD    (AAA Similarity)

Therefore,
ar(AOB)/ar(COD) = (AB/CD)²
Now, putting the value of AB.
ar(AOB)/ar(COD) = (2CD/CD)²
ar(AOB)/ar(COD) = 4/1
Required ratio = 4:1 Ans.

Q.No.3. In Fig. 6.46 ABC and DBC are two triangles on the same base BC. If AD intersect BC at O, show that
ar(ABC)/ar(DBC) = AO/DO

Solution:

Draw perpendicular AP and DP on BC.
Consider in triangles APO and DQO.
<APO = <DQO (Each 90⁰)
<AOP = <DOQ  (Vertically opposite angles)

△APO ∼ △DQO
Therefore,
ar(APO)/ar(DQO) = (AP/DQ)² = (AO/DO)² = (PO/QO)² -------------- (i)

Again,
ar(ABC) = BCxAP/2
ar(DBC) = BCxDQ/2
ar(ABC)/ar(DBC) = AP/DQ

Hence from (i),
ar(ABC) = ar(DBC) = AO/DO Proved.

Q.No. 4. If the areas of two similar triangles are equal, prove that they are congruent.

Solution:

Consider about two triangles ABC and PQR.
Therefore,
ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)² ------------------ (i)

Again,
ar(ABC) = ar(PQR)       (Given)        ----------------------  (ii)

Now, from (i) and (ii).

AB = PQ
BC = QR
AC = PR

Hence,
△ABC≅△PQR Proved.

Q.No. 5. D, E and F respectively the mid points of the sides AB, BC and CA of triangle ABC. Find the ratio of the areas of △ DEF and △ABC.

Solution:

D and F are the mid points of AB and CA.
Therefore,
AD/DB = AF/FC
So,
DF ॥ BC
and
DF = BC/2 ------------- (i)

Similarly,
DE = AC/2 ------------- (ii)
and
EF = AB/2 ----------- -- (iii)

Now, from (i), (ii) and (iii).
ADEF, BEFD and CFDE are parallelograms.

Now, consider in triangles ABC and DEF.
<A = <E ---------------- (ADEF is a parallelogram)
<B = <F ---------------- (BEFD is a parallelogram)
<C = <D ----------------- ( CFDE is parallelogram)

Hence,
△ABC ∼ △EFD
ar(ABC)/ar(DEF) = (AB/EF)²
Now, putting the value of EF.
ar(ABC)/ar(DEF) = 4:1

Therefore,
ar(DEF)/ar(ABC) = 1:4 Ans.


Q.No. 6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution:

Consider about two similar triangles ABC and PQR.
AX, BY and CZ are the medians of ABC. PK, QL and RM are medians of PQR.

Since,
ABC and PQR are similar triangles.
ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)² ---------------------- (i)

Since,
AX and PK are medians of ABC and PQR.
Therefore,
Triangles ABX and PQK are similar.
ar(ABX)/ar(PQK) = (AB/PQ)² = (AX/PK)² ---------------------- (ii)

Now, from (i) and (ii).

The ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians. Proved.

7. Prove that area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution:

Let the side of the square be X.
Diagonals of the square = ✔️2 X

Area of equilateral triangle on diagonal = 2✔️3X²/4 ------- (i)
Area of equilateral triangles on side = ✔️3X²/4 ------ (ii)

Now, from (i) and (ii)

Area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Proved

Areas of Similar Triangles

Class X

Theorem

"The ratios of areas of two triangles is equal to the square of the ratio of their corresponding sides".

Solution:-

Given

△ABC ~ △PQR

Therefore,

AB/PQ = BC/QR = CA/RP

To prove that

ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)²

Construction

Draw AM ⫡ BC and PN ⫡ QR

Proof

ar(ABC) = BCxAM/2 ---------------- (i)

ar(PQR) = QRxPN/2 ----------------- (ii)

Now, from (i) and (ii)

ar(ABC)/ar(PQR) = BCxAM/QRxPN ------------ (iii)

Now, consider in △ABM and △PQN

<B = <Q     (△ABC ~ △PQR)

<M = <N    ( Each 90⁰)

Hence,

△ABM ~ △PQN  (AA similarity)

Therefore,

AM/PN = AB/PQ

Hence,

AM/PN = AB/PQ = BC/QR = CA/RP

Now, from (iii)

ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)² Proved.