Chapter 20: Motion Of Projectile¶

import math

#Initilization of variables

v_o=500 # m/s # velocity of the projectile
alpha=30 # angle at which the projectile is fired
t=30 # seconds
g=9.81 # m/s^2 # acc due to gravity

# Calculations

v_x=v_o*cos(alpha*(pi/180)) # m/s # Initial velocity in the horizontal direction
v_y=v_o*sin(alpha*(pi/180)) # m/s # Initial velocity in the vertical direction

# MOTION IN HORIZONTA DIRECTION:
V_x=v_x # m/s # V_x=Horizontal velocity after 30 seconds

# MOTION IN VERTICAL DIRECTION: # using the eq'n v=u+a*t
V_y=v_y-(g*t) # m/s # -ve sign denotes downward motion

# Let the Resultant velocity be v_R. It is given as,
v_R=((V_x)**2+(-V_y)**2)**0.5# m/s
theta=arctan((-V_y)/V_x)*(180/pi) # degree # direction of the projectile

# Results

print"The velocity of the projectile is ",round(v_R,2),"m/s" # The answer of velocity is wrong in the text book.
print"The direction of the projectile is ",round(theta,2),"degree" # -ve value of theta indicates that the direction is in downward direction

import math

# Initilization of variables

v_A=10 # m/s # velocity of body A
alpha_A=60 # degree # direction of body A
alpha_B=45 # degree # direction of body B

# Calculations

# (a) The velocity (v_B) for the same range is given by eq'n;
v_B=((v_A**2*sin(2*alpha_A*(pi/180)))/(sin(2*alpha_B*(pi/180))))**0.5 # m/s

# (b) Now velocity v_B for the same maximum height is given as,
v_b=((v_A**2)*((sin(alpha_A*(pi/180)))**2/(sin(alpha_B*(pi/180)))**2))**0.5 # m/s

# (c) Now the velocity (v) for the equal time of flight is;
v=(v_A*sin(alpha_A*(pi/180)))/(sin(alpha_B*(pi/180))) # m/s

# Results

print"(a) The velocity of body B for horizontal range is ",round(v_B,1),"m/s"
print"(b) The velocity of body B for the maximum height is ",round(v_b,2),"m/s"
print"(c) The velocity of body B for equal time of flight is ",round(v,2),"m/s"

import math

# Initilization of variables

y=3.6 # m # height of the wall
x_1=4.8 # m # position of the boy w.r.t the wall
x_2=3.6 # m # distance from the wall where the ball hits the ground
g=9.81 # m/s^2 # acc due to gravity

# Calculations

# The range of the projectile is r, given as,
r=x_1+x_2 # m

# Let the angle of the projection be alpha, which is derived and given as,
alpha=arctan((y)/(x_1-(x_1**2/r)))*(180/pi) # degree

# Now substuting the value of alpha in eq'n 3 we get the least velocity (v_o) as;
v_o=((g*r)/(sin(2*alpha*(pi/180))))**0.5 # m/s

# Results

print"The least velocity with which the ball can be thrown is ",round(v_o,2),"m/s"
print"The angle of projection for the same is ",round(alpha,1),"degree"

import math

# Initilization of variables

v_o=400 # m/s # initial velocity of each gun
r=5000 # m # range of each of the guns
g=9.81 # m/s^2 # acc due to gravity
p=180 # degree 
pi=3.14

# Calculations

# now from eq'n 1
theta_1=(arcsin(r*g/v_o**2)*(180/pi))/2 # degree # angle at which the 1st gun is fired

# from eq'n 3
theta_2=(p-2*theta_1)/2 # degree 

# For 1st & 2nd gun, s is
s=r # m

# For 1st gun 
v_x=v_o*cos(theta_1*(pi/180)) # m/s

# Now the time of flight for 1st gun is t_1, which is given by relation,
t_1=s*(v_x)**-1 # seconds

# For 2nd gun
V_x=v_o*cos(theta_2*(pi/180))

# Now the time of flight for 2nd gun is t_2
t_2=s/V_x # seconds

# Let the time difference between the two hits be delta.T. Then,
T=t_2-t_1 # seconds

# Results

print"The time difference between the two hits is ",round(T,2),"seconds" #answer varies by 0.6 sec due to decimal variance

import math

# Initilization of variables

h=2000 # m/ height of the plane
v=540*1000*3600**-1 # m/s # velocity of the plane
g=9.81 # m/s^2 # acc due to gravity
pi=3.14

# Calculations

# Time t required to travel down a height 2000 m is given by eq'n,
u=0 # m/s # initial velocity
t=(2*h/g)**0.5 # seconds

# Now let s be the horizonta distance travelled by the bomb in time t seconds, then
s= v*t # m

# angle is given as theta,
theta=arctan(h/s)*(180/pi) # degree

# Results

print"The pilot should release the bomb from a distance of ",round(s),"m" #the answer varies by 1m due to decimal variance
print"The angle at which the target would appear is ",round(theta,1),"degree"

import math

# Initilization of variables

theta=30 # degree # angle at which the bullet is fired
s=-50 # position of target below hill
v=100 # m/s # velocity at which the bullet if fired
g=9.81 # m/s^2 

# Calculations

v_x=v*cos(theta*(pi/180)) # m/s # Initial velocity in horizontal direction
v_y=v*sin(theta*(pi/180)) # m/s # Initial velocity in vertical direction

# (a) Max height attained by the bullet
h=v_y**2/(2*g) # m

# (b)Let the vertical Velocity with which the bullet will hit the target be V_y. Then,
V_y=((2*-9.81*s)+(v_y)**2)**0.5 # m/s # the value of V_y is +ve & -ve

# Let V be the velocity with wich it hits the target
V=((v_x)**2+(V_y)**2)**0.5 # m/s

# (c) The time required to hit the target
a=g # m/s^2
t=(v_y-(-V_y))/a # seconds

# Results

print"(a) The maximum height to which the bullet will rise above the soldier is ",round(h,1),"m"
print"(b) The velocity with which the bullet will hit the target is ",round(V,1),"m/s"
print"(c) The time required to hit the target is ",round(t,1),"seconds"

import math

# Initilization of variables

W=30 # N # Weight of the hammer
theta=30 # degree # ref fig.20.12
mu=0.18 # coefficient of friction
s=10 # m # distance travelled by the hammer # fig 20.12
g=9.81 # m/s^2 # acc due to gravity

# Calculations

# The acceleration of the hammer is given as,
a=g*((sin(theta*(pi/180)))-(mu*cos(theta*(pi/180)))) # m/s^2

# The velocity of the hammer at point B is,
v=(2*a*s)**0.5 # m/s

# Let the initial velocity of the hammer in horizontal direction be v_x & v_y in vertical direction, Then,
v_x=v*cos(theta*(pi/180)) # m/s
v_y=v*sin(theta*(pi/180)) # m/s

# MOTION IN VERTICAL DIRECTION
# Now, let time required to travel vertical distance (i.e BB'=S=5 m) is given by finding the roots of the second degree eq'n as,
# From the eq'n 4.9*t^2+4.1*t-5=0,
a=4.9
b=4.1
c=-5

# The roots of the eq'n are,
t=((-b)+((b**2-(4*a*c))**0.5))/(2*a)

# MOTION IN HORIZONTAL DIRECTION
# Let the horizotal distance travelled by the hammer in time t be s_x.Then,
s_x=v_x*cos(theta*(pi/180))*t # m
x=1+s_x # m

# Results

print"The distance x where the hammer hits the round is ",round(x,2),"m"

import math

# Initilization of variables
s=1000 # m # distance OB (ref fig.20.13)
h=19.6 # m # height of shell from ground
g=9.81 # m/s^2 # acc due to gravity

# Calculations

# MOTION OF ENTIRE SHELL FROM O to A.
v_y=(2*(g)*h)**0.5 # m/s # initial velocity of shell in vertical direction
t=v_y/g # seconds # time taken by the entire shell to reach point A
v_x=s/t # m/s # velocity of shell in vertical direction

# VELOCITIES OF THE TWO PARTS OF THE SHELL AFTER BURSTING AT A:

# Let v_x2 be the horizontal velocity of 1st & the 2nd part after bursting which is given as,
v_x2=v_x*2 # m/s

# Now distance BC travelled by part 2 is
BC=v_x2*t # m

# Distance from firing point OC
OC=s+BC # m

# Results

print"(a) The velocity of shell just before bursting is ",round(v_x),"m/s"
print"(b) The velocity of first part immediately before the shell burst is ",round(v_x),"m/s"
print"(c) The velocity of second part immediately after the shell burst is ",round(v_x2,1),"m/s"
print"(b) The distance between the firing point & the point where the second part of the shell hit the ground is ",round(OC),"m"

import math

# Initilization of variables

v_o=200 # m/s # initial velocity
theta=60 # degree # angle of the incline
y=5 # rise of incline
x=12 # length of incline
g=9.81 # m/s^2 # acc due to gravity


# Calculations

# The angle of the inclined plane with respect to horizontal
beta=arctan(y*x**-1)*(180/pi) # degree

# The angle of projection with respect to horizontal
alpha=90-theta # degree

# Range is given by eq'n (ref. fig.20.14)
AB=(2*v_o**2*(sin((alpha-beta)*(pi/180)))*cos(alpha*(pi/180)))/(g*(cos(beta*(pi/180)))**2) # m

# Range AC when the short is fired down the plane
AC=(2*v_o**2*(sin((alpha+beta)*(pi/180)))*cos(alpha*(pi/180)))/(g*(cos(beta*(pi/180)))**2) # m
BC=AB+AC # m

# Results

print"The range covered (i.e BC) is ",round(BC),"m" #due to decimal variancce answer varies by 1m