Suppose you are a general in the Napoleonic wars. You are no top of a plain 50 m high overlooking the enemy soldiers. You now have a brand new cannon that will decimate the enemy. Unfortunately in the christening process you used some rather cheap grade champagne and all the water in the champagne rusted the hinges of the cannon so that it is stuck angled upward at an angle of 30 from the horizontal. Your trusty new cannon expert insisted that the ball will reach a maximum height of 150m from the ground but unfortunately he was killed before he could tell you where the cannonball would land on the plain below. You have soldiers on the ground and you want to make sure that only the enemy soldiers are killed by your uber cannon. Determine the range of the cannon in order to keep your soldiers safe.
Y^2 = Yo^2 + 2g*h = o
Yo^2 - 19.6*(150-50) = 0
Yo^2 = 1960
Yo = 44.27 m/s. = Ver. component of the
initial velocity.
Vo = Yo/sin A = 44.27/sin30 = 88.54 m/s.
= Initial velocity.
Xo = Vo*cos A = 88.54*cos30 = 76.68 m/s.
= Hor. component of initial velocity.
Y = Yo + g*Tr = 0 @ max. ht.
44.27 - 9.8*Tr = 0
9.8Tr = 44.27
Tr = 4.52 s. = Rise time.
h = 0.5g*t^2 = 150 m
4.9t^2 = 150
t^2 = 30.61
Tf = 5.53 s. = Fall time.
Range = Xo*(Tr+Tf) = 76.68*(4.52+5.53)=
770.6 m.
To determine the range of the cannon and ensure the safety of your soldiers, we need to find out where the cannonball will land on the plain below.
One way to approach this is by considering the motion of the cannonball as a projectile. We can make use of the kinematic equations of motion to solve for the range.
First, let's break down the initial velocity of the cannonball. Since the cannon is stuck angled upward at an angle of 30 degrees from the horizontal, we can split the velocity into horizontal and vertical components.
The vertical component of the initial velocity can be found using trigonometry:
Vertical component = initial velocity * sin(angle)
Vertical component = V * sin(30)
The horizontal component of the initial velocity can also be found using trigonometry:
Horizontal component = initial velocity * cos(angle)
Horizontal component = V * cos(30)
Now, let's focus on the vertical motion of the cannonball. We know that it will reach a maximum height of 150m above the ground. At the maximum height, the vertical component of the velocity will become zero.
Using the kinematic equation:
Vertical component^2 = initial velocity^2 - 2 * acceleration * height
(0)^2 = (V * sin(30))^2 - 2 * (-9.8) * 150
Simplifying the equation gives:
0 = (V^2 * sin^2(30)) + 2940
Now, let's focus on the horizontal motion of the cannonball. The horizontal component of velocity remains constant throughout the motion.
Using the kinematic equation:
Range = horizontal component * time
Range = (V * cos(30)) * time
To find the time, we can use the vertical component of velocity:
Vertical component = initial velocity * sin(angle) - acceleration * time
0 = V * sin(30) - 9.8 * time
Simplifying the equation gives:
time = V * sin(30) / 9.8
Substituting the time back into the range equation:
Range = (V * cos(30)) * (V * sin(30) / 9.8)
Range = (V^2 * (1/2)) / 9.8
Now, we can combine the equations for the range and the equation for the vertical motion:
(V^2 * (1/2)) / 9.8 = (V^2 * sin^2(30)) + 2940
We can solve this equation for V (initial velocity), and then calculate the range using the formula we derived earlier:
Range = (V * cos(30)) * (V * sin(30) / 9.8)
By solving this equation, we can find the range of the cannon and determine the safe zone for your soldiers.