As everyone knows, the 16th century French people always first tested their cano

Question

As everyone knows, the 16th century French people always first tested their canons in a vacuum, so as to properly calibrate them in the simplest of scenarios. As a first test, a canon sitting on the Earth's surface fires a canon ball vertically at a starting velocity of 200 m/s. The canon ball weighs 5 kg. and since the canon is sitting in a vacuum, there is no air resistance. Given the initial conditions, find particular solutions for both the position and the velocity of the canon ball. How high does the canon ball fly? How much time does it take to make it to this apex? Now satisfied with the canon's calibration, the French turn it towards their arch-enemies the British. This time, the canon fires at an angle of 45 degrees. The initial velocity is 200 m/s, and the air resistance is defined here as R = -02v. Here, we are therefore dealing with two spatial dimensions (the horizontal and vertical), and thus the velocity of the canon ball is not a 1 dimensional vector. The canon ball once again weighs 5 kg. You'll note in this case that both spatial dimensions have unique force diagrams. Find particular solutions for both j(j) and y(t) given the initial conditions. 2 What is the maximum height the canon ball attains? At what time does this happen? If the British camps are. at a minimum, 1.2 km away, does the canon ball manage to win it for the French, or do the British live to fight another day?

Explanation / Answer

4]a] cannon is firing vertically upward

m=5 kg

Viy=200 m/s

ay=-9.8 m/s2

Let y be the vertical distance covered by the ball and t be the time,Using the kinematic equation

y=Viy* t+ 1/2ayt2

y=200t-4.9t2--------------------------------------this equation gives the vertical height of cannon ball for time t.

Let Vfy is the final velocity and can be computed by the initial condition as

Vfy= Viy+ayt

4]b]

At apex, Vfy=0 m/s

Hence we have the kinematic equation Vfy2= Viy2+2ayy in the form

02=2002+2[-9.8]* y

-2002=-19.6y

y= 2040.8 m

The vertical height of the cannon ball= 2040.8 m

Time taken for this height is, Vfy= Viy+ay*t

0=200-9.8 t

200=9.8 t

Time taken for this height is,t=20.41 seconds

Next case- cannon is firing with an angle of 45 o

a] so here 2D motion[projectile one]

Along the x axis , Vx remains constant

Vx= Vi*cos 450= 200*cos 450= 141.42 m/s

x[t]= Vx* t--------------------------equation 1

Along the y axis , Vy varies; ay=-9.8 m/s2

Viy= Vi*cos 450= 200*cos 450= 141.42 m/s

Viy= 141.42 m/s

Vfy= Viy+ayt=141.42-9.8t

Vfy=141.42-9.8t

y[t]=Viy* t+ 1/2ayt2 =141.42t-4.9t2

y[t]=141.42t-4.9t2--------------------------equation 2

{equation 1 and 2 gives the solution for x[t] and y[t]}

b]

At maximum height, Vfy=0 m/s

Vfy=141.42-9.8t

0-141.42=-9.8 t

t=141.42/9.8=14.43 seconds

At maximum height , time =14.43 seconds

y[t]=Viy* t+ 1/2ayt2

   =141.42t-4.9t2

   =[141.42*14.43] -[4.9*14.432]

= 2040.78-1020.3=1020.5m

Maximum height the cannon ball attains is 1020.5m

c]

Horizontal displacement made by the cannon ball is ; x[t]= Vx* t=141.42 m/s*14.43s= 2040.7 m=2.041km

The british camp is 1.2 km away

so here cannon ball manage to win it for the french

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