# Chapter 14:Projectiles¶

## Example 14.1,Page No.507¶

In [1]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables
alpha=60 #Degrees
R=5000 #m #Horizontal Range
g=9.81 #m/s**2 #Acceleration due to gravity

#Calculation

#By equation of Horizontal Range,we get
u=((R*g)*(sin(2*alpha*pi*180**-1))**-1)**0.5 #m/s

#Max Height attained by projectile
h_max=u**2*(sin(alpha*pi*180**-1))**2*(2*g)**-1

#Result
print"Velocity of projection is",round(u,2),"m/s"
print"Max Height attained by projectile is",round(h_max,2),"m"

Velocity of projection is 237.99 m/s
Max Height attained by projectile is 2165.06 m


## Example 14.2,Page No.507¶

In [2]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables
u=100 #m/s #Velocity
alpha=30 #Degrees #Angle made by projectile with horizontal
g=9.81 #m/s #Acceleration due to gravity

#Calculation

#Horizontal Range
R=u**2*sin(2*alpha*pi*180**-1)*g**-1 #m

#MAx height attained
h_max=u**2*sin(alpha*pi*180**-1)*g**-1 #m

#Time of flight
T=2*u*sin(alpha*pi*180**-1)*g**-1 #s

#Result
print"Horizontal Range is",round(R,2),"m"
print"MAx Height attained is",round(h_max,2),"m"
print"Time of Flight is",round(T,2),"s"

Horizontal Range is 882.8 m
MAx Height attained is 509.68 m
Time of Flight is 10.19 s


## Example 14.3,Page No.508¶

In [3]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables
#R=4*h_max

#Calculation

#Equation ofhorizontal Range is
#R=u**2*sin(2*alpha)*g**-1   .........1

#Equation of  MAx height
#h_max=u**2*sin(alpha)**2*(2*g)**-1    .........2

#After simplifying both equations,we get
alpha=np.arctan(1)*(pi**-1*180)

#Result
print"Angle of projections is",round(alpha,2),"degrees"

Angle of projections is 45.0 degrees


## Example 14.4,Page No.508¶

In [2]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables
u=20 #m/s #Velocity
x=20 #m #X cordinate of trajectory
y=8 #m #Y cordinate of trajectory

#Calculation

#equation of trajectory
#y=x*tan(alpha)-g*x**2*(2*u**2*cos(alpha)**2)**-1
#After substituting values and further simplifying we get quadratic equation
#4.905*(tan(alpha))**2-20*tan(alpha)+12.905=0
a=4.905
b=-20
c=12.905

X=b**2-4*a*c

y1=(-b+X**0.5)*(2*a)**-1
y2=(-b-X**0.5)*(2*a)**-1

alpha1=np.arctan(y1)*(pi**-1*180) #Degrees
alpha2=np.arctan(y2)*(pi**-1*180) #Degrees

#Result
print"Angle of projection of particle is:alpha1",round(alpha1,2),"Degrees"
print"                                  :alpha2",round(alpha2,2),"Degrees"

Angle of projection of particle is:alpha1 73.01 Degrees
:alpha2 38.79 Degrees


## Example 14.5,Page No.509¶

In [5]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

x=4.8 #m #X-cord of projectile
y=3.6 #m #y-cord of projectile
alpha=60 #Degrees #Inclination of jet
g=9.81 #m/s**2

#Calculation

#Equation of trajectory
#y=x*tan(alpha)-g*x**2*(2*u**2*cos(alpha)**2)**-1
#After further simplifying we get
u=((g*x**2)*(2*(cos(alpha*pi*180**-1))**2*((x*tan(alpha*pi*180**-1))-y))**-1)**0.5 #m/s

#Result
print"required velocity of jet at nozzle exit is",round(u,2),"m/s"

required velocity of jet at nozzle exit is 9.79 m/s


## Example 14.6,Page No.510¶

In [6]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables

u=250 #m/s #Velocity
x=4000 #m #x-cord
y=700 #m #y-cord

#Calculation

#equation of trajectory
#y=x*tan(alpha)-g*x**2*(2*u**2*cos(alpha)**2)**-1
#After substituting values and further simplifying we get quadratic equation
#1255.68*tan(alpha)**2-4000*tan(alpha)+1955.68=0
a=1255.68
b=-4000
c=1955.68

X=b**2-4*a*c

y1=(-b+X**0.5)*(2*a)**-1
y2=(-b-X**0.5)*(2*a)**-1

alpha1=np.arctan(y1)*(pi**-1*180) #Degrees
alpha2=np.arctan(y2)*(pi**-1*180) #Degrees

#Result
print"Firing angle to hit the target is:alpha1",round(alpha1,2),"Degrees"
print"                                 :alpha2",round(alpha2,2),"Degrees"

Firing angle to hit the target is:alpha1 68.83 Degrees
:alpha2 31.09 Degrees


## Example 14.7,Page No.511¶

In [7]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables

alpha1=15 #Degrees #Angle of projectile 1
alpha2=45 #Degrees #Angle of projectile2
g=9.81 #m/s**2
#R1=R-12
#R2=R+24

#Calculation

#form Equation of horizontal Range we get,
#R-12=u**2*sin(2*alpha1)*g**-1  ..................1

#R+24=u**2*sin(2*alpha2)*g**-1   ....................2

#Dividing equation 1 by 2 and further simplifying we get
R=24+24 #m

#Sub value of R in equation 2 we get
#u**2=g*72

#Sub values of R and u**2 in equation of horizontal range we get
#sin(2*alpha)=R*g*(g*72)**-1
#LEt sin(2*alpha)=X
X=R*g*(g*72)**-1
alpha=(np.arcsin(X)*(180*pi**-1))*2**-1

#Result
print"Angle of projection to hit the mark is",round(alpha,2)

Angle of projection to hit the mark is 20.91


## Example 14.8,Page No.512¶

In [8]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables

u=100 #m/s #initial Velocity
alpha=30 #DEgrees #Angle of projection
g=9.81 #m/s**2
h=80 #m #Height below B

#Calculation

#Max Height
H=u**2*(sin(alpha*pi*180**-1))**2*(2*g)**-1 #m

#Vertical Distance
S=H+h #m

#Vertical Component of velocity striking the target
v2=(2*g*S)**0.5 #m/s

#Horizontal component of velocity
v=u*cos(alpha*pi*180**-1)

#Actual Velocity with which bullet strikes the target
V=(v2**2+v**2)**0.5

#Angle made by actual velocity striking velocity
theta=np.arctan(v2*v**-1)*(pi**-1*180)

#Result
print"Max height attained by bullet",round(H,2),"m"
print"Actual Velocity with which it will strike the target",round(theta,2),"Degrees"

Max height attained by bullet 127.42 m
Actual Velocity with which it will strike the target 36.38 Degrees


## Example 14.9,Page No.514¶

In [3]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables
h=150 #m #Height of cliff
u=180 #m/s #Initial Velocity
alpha=30 #ANgle of projection
g=9.81

#Calculation

#From equation of vertical Distance
#y=u*sin(Alpha)*t-0.5*g*t**2
#Sub values and further simplifying we get
#4.905t**2-90*t-150=0
a=4.905
b=-90
c=-150

X=(b**2-4*a*c)**0.5

#Total time of flight
t=(-b+(X))*(2*a)**-1

#Horizontal Distance
h1=u*cos(alpha*pi*180**-1)*t

#MAx Height
H=u**2*sin(alpha*pi*180**-1)**2*(2*g)**-1 #m

#ELEvation above the ground
H2=h+H #m

#Result
print"Horizontal Distance from gun is",round(h1,2),"m"
print"Elevation above the ground",round(H2,2),"m"

Horizontal Distance from gun is 3099.98 m
Elevation above the ground 562.84 m


## Example 14.10,Page No.515¶

In [10]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables

h=6 #m #Height of tunnel
u=50 #m/s #Initial Velocity
g=9.81

#Calculation

alpha=np.arcsin(((h*2*g)*((u**2)**-1))**0.5)*(pi**-1*180)

#Horizontal Range
R=u**2*sin(2*alpha*pi*180**-1)*g**-1

#Result
print"Angle of projection is",round(alpha,2),"DEgrees"
print"Horizontal Range is",round(R,2),"m"

Angle of projection is 12.53 DEgrees
Horizontal Range is 107.96 m


## Example 14.11,Page No.516¶

In [11]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables
alpha1=30 #Degrees #Angle of projection1
alpha2=30 #Degrees #Angle of projection2
u1=350 #m/s #Velocity of projection at A
u2=300 #m/s #Velocity of projection at B
L_AB=30 #m #distance between A nd B
g=9.81 #m/s**2

#Calculation

#Horizontal component of velocity at A
V_A=u1*cos(alpha1*pi*180) #m/s

#Horizontal component of velocity at B
V_B=u2*cos(alpha2*pi*180) #m/s

#Horizontal Distance covered by shot A
#x=(u1*cos(alpha1)*t   ..........................1

#Horizontal Distance covered by shot B
#15*(3)**0.5-x=(u2*cos(alpha2))*t   ................2

#Adding Equations 1 and 2 we get
t=15*(3)**0.5*((u1+u2)*cos(alpha1*pi*180**-1))**-1

#sub values in equation 1 we get
x=(u1*cos(alpha1*pi*180**-1))*t

#Intial velocity shot in vertical direction
V2=u1*sin(alpha1*pi*180**-1) #m/s

#Distance in vertical direction
y=V2*t-g*t**2*2**-1

#Result
print"Time when these two shots meet",round(t,2),"s"
print"Vertical Distance at which they meet",round(y,3),"m"

Time when these two shots meet 0.05 s
Vertical Distance at which they meet 8.066 m


## Example 14.12,Page No.518¶

In [4]:
import math

#Initilization of Variables
v=720*10**3*3600**-1 #m/s #Speed of air craft
u=200 #m/s #Velocity of bomb
H=1000 #m #Height
g=9.81

#Calculation

#Time requaired by bomb
t=(H*2*g**-1)**0.5 #s

#Horizontal distance of air craft
h=u*round(t,3) #m

#Result
print"Time requaired by bomb to reach the ground",round(t,2),"s"
print"Horizontal distance of air craft",round(h,2),"m"

Time requaired by bomb to reach the ground 14.28 s
Horizontal distance of air craft 2855.6 m


## Example 14.13,Page No.519¶

In [13]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables

v=108*10**3*3600**-1 #m/s #Spedd of air craft
u=30 #m/s #Horizontal Velocity
H=1000 #m #Height
g=9.81 #m/s**2

#Calculation

#Time taken by bomb
t=(H*2*g**-1)**0.5 #s

#Horizontal Distance OB
h=u*t #m

#Velocity of bomb hitting the ground
V=g*t #m/s**2

#Resultant Velocity
V2=(u**2+V**2)**0.5 #m/s

#Direction in which the bomb hits the ground
theta=np.arctan(V*u**-1)*(180*pi**-1)

#Result
print"HOrizontal Distance of air craft from target",round(h,2),"m"
print"Velocity of bomb",round(V2,2),"m/s"
print"Direction of bomb",round(theta,2),"Degrees"

HOrizontal Distance of air craft from target 428.35 m
Velocity of bomb 143.25 m/s
Direction of bomb 77.91 Degrees


## Example 14.14,Page No.520¶

In [14]:
import math

#Initilization of Variables
H=2000 #m #Height of aeroplane
u=10**6*3600**-1 #Velocity of plane #m/s
g=9.8 #m/s**2

#Calculation

#Time required by bomb
t=(2*H*g**-1)**0.5 #s

#Horizontal Distance travelled by bomb
h=u*t #m

#Result
print"Time required by bomb to reach the ground",round(t,2),"s"
print"Horizontal Distance travelled by bomb",round(h,2),"m"

Time required by bomb to reach the ground 20.2 s
Horizontal Distance travelled by bomb 5611.96 m


## Example 14.16,Page No.521¶

In [15]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables
L_AB1=4 #m #LEngth of AB in Horizontal Distance
L_AB2=2 #m #LEngth of AB in Vertical Distance
g=9.81 #m/s**2

#Calculation

#time
t=((L_AB2*2)*(9.81)**-1)**0.5

#Minimum veloacity of motorcyclye at A in Horizontal Direction
u=L_AB1*t**-1 #m/s

#Vertical Component of velocity
v=g*t #m/s

#Resultant velocity at B
V=(u**2+v**2)**0.5*3600*(10**3)**-1 #km/hr

#Inclination
theta=np.arctan(v*u**-1)*(180*pi**-1)

#Result
print"Inclination of motorcycle clearing the ditch",round(theta,2),"Degrees"
print"MAgnitude of velocity of motorcycleafter clearing ditch",round(V,2),"km/hr"

Inclination of motorcycle clearing the ditch 45.0 Degrees
MAgnitude of velocity of motorcycleafter clearing ditch 31.89 km/hr


## Example 14.17,Page No.522¶

In [5]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables
alpha=25 #Degrees #Angle of ramp
L_BC1=4 #m #Horizontal Distance between B and C
L_BC2=2 #m #vertical Distance between B and C
g=9.81 #m/s**2

#Calculation

##From equation of path
X=(L_BC2+L_BC1*tan(alpha*pi*180**-1))
Y=(g*L_BC1**2)*(2*(cos(alpha*pi*180**-1))**2)**-1
u=(X**-1*Y)**0.5

#Result
print"Minimum speed of motorcycle is",round(u,2),"m/s"

Minimum speed of motorcycle is 4.97 m/s


## Example 14.18,Page No.526¶

In [6]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables
u=40 #m/s #Velocity of projection
alpha=50 #Degrees #Angle of projection
beta=20 #degrees #Ang;e if incline plane
g=9.81

#Calculation

#Time of flight
T=2*u*sin((alpha-beta)*pi*180**-1)*(g*cos(beta*pi*180**-1))**-1

#Range
R=2*u**2*cos(alpha*pi*180**-1)*sin((alpha-beta)*pi*180**-1)*(g*(cos(beta*pi*180**-1))**2)**-1

#Let alpha2 be the angle of projection for max range
alpha2=(90+beta)*2**-1

#MAx range up the plane
R2=u**2*(g*(1+sin(beta*pi*180**-1)))**-1

#Result
print"Time of flight is",round(T,2),"s"
print"range up the plane is",round(R,2),"m"
print"Max range up the plane",round(R2,2),"m"

Time of flight is 4.34 s
range up the plane is 118.73 m
Max range up the plane 121.53 m


## Example 14.19,Page No.527¶

In [7]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

u=30 #m/s #velocity
alpha=55 #angle of projection
beta=20 #angle of plane
g=9.81

#Calculation
#Maximum Range
R=u**2*(g*(1+sin(beta*pi*180**-1)))**-1 #m

#Time of Flight
T=2*u*sin((alpha-beta)*pi*180**-1)*(g*cos(beta*pi*180**-1))**-1

#Result
print"Time of Fliight is",round(T,2),"s"
print"Maximum Range is ",round(R,2),"m"

Time of Fliight is 3.73 s
Maximum Range is  68.36 m


## Example 14.20,Page No.527¶

In [12]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

u=3 #m/s #Velocity of projection
alpha=25 #Degrees #Angle of projection
beta=25 #degrees #Angle of plane with horizontal
g=9.81 #m/s**2

#Calculation

#Range
R=2*u**2*cos(alpha*pi*180**-1)*sin((alpha+beta)*pi*180**-1)*(g*(cos(beta*pi*180**-1))**2)**-1

#L_BC=y
#L_AC=x

#From tan(beta) we get
#y=0.466*x   .................(1)

#From Equation (L_AB)**2=(L_BC**2+L_AC)**2
#After sub values and further simplifying we get
x=(2.4*(1.217)**-1)**0.5 #m

#Sub in equation 1 we get
y=0.466*x

#Result
print"Co-ordinates of point B are:x",round(x,2),"m"
print"                           :y",round(y,2),"m"

Co-ordinates of point B are:x 1.4 m
:y 0.65 m


## Example 14.21,Page No.528¶

In [9]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables

L_AB=75 #m #LEngth of AB
h=19.6 #m #Height of point of release
beta=26.564 #Degrees #angle of plane

#Calculation

#Length AC
L_AC=L_AB*cos(beta*pi*180**-1) #m

#From horizontal Distance
#u*cos(alpha)=L_AC*t**-1   .............1

#Vertical motion motion from point of release
#u*sin(alpha)=19.6   ....................2

#vertical Distance
y=L_AB*sin(beta*pi*180**-1) #m

#Vertical distance from point of release travelled by ball
#y=u*sin(alpha)*t-0.5*g*t**2

#sub value of y in above equation and further simplifyin we get
#t**2-4t-6.84=0
a=1
b=-4
c=-6.84

X=b**2-4*a*c

t=(-b+X**0.5)*(2*a)**-1

#sub value in equation 1 we get
#Let ucos(alpha)=X
X=67.08*t**-1  #..........................3

#Dividing equation 1 by 3 we get
alpha=np.arctan(19.6*12.68**-1)*(180*pi**-1)

#sub in equation 1 we get
u=12.68*cos(alpha*pi*180**-1)**-1

#Result
print"Initial Velocity",round(u,2),"m/s"
print"Inclination is",round(alpha,4),"Degrees"

Initial Velocity 23.34 m/s
Inclination is 57.0996 Degrees