Expert: Paul Klarreich - 12/7/2006

Question
Hello, I'm a first year college student and I was wondering if you could help me with this problem, it's somewhat confusing...

At time t=0, a cannon sitting on level ground fires a shell with muzzle speed 400ft/sec, directed at an angle of pi/6 upward from the horizontal. Let us introduce a coordinate system with the origin at the position of the cannon, the y axis vertical, and the x axis along hte ground, pointing in the direction in which the cannon fires.

a) The initial velocity of the shell in the x direction is 400cos(pi/6) ft/sec & the y direction is 400sin(pi/6) ft/sec.
b) The x coordinate of the shell's position, x(t) satisfies x"(t) = 0, since no forces acts in the x direction. (This is true until the shell hits the ground; we are ignoring air resistance.) Using a, find x(t).
c) The y coordinate of the shell's position, y(t), satisfies y"(t) = -32; this is the acceleration due ot gravity (again, until the shell hits the ground.) Find y(t).
d. Find the Initial Time (t sub 0) that the shell hits the ground, and the x position x(t sub 0) at that time (This position is called the range).

So far, I think I have figured out C, which is y'(t)= -32t+C, C = 200. Other than that, I am lost! Thanks alot in advance!!!

Answer
Questioner:  Faizan
Category:  Calculus
 
Subject:  Physics-related Calculus
Question:  Hello, I'm a first year college student and I was wondering if you could help me with this problem, it's somewhat confusing...

At time t=0, a cannon sitting on level ground fires a shell with muzzle speed 400ft/sec, directed at an angle of pi/6 upward from the horizontal. Let us introduce a coordinate system with the origin at the position of the cannon, the y axis vertical, and the x axis along hte ground, pointing in the direction in which the cannon fires.

a) The initial velocity of the shell in the x direction is 400cos(pi/6) ft/sec & the y direction is 400 sin(pi/6) ft/sec.

b) The x coordinate of the shell's position, x(t) satisfies x"(t) = 0, since no forces acts in the x direction. (This is true until the shell hits the ground; we are ignoring air resistance.) Using a, find x(t).

c) The y coordinate of the shell's position, y(t), satisfies y"(t) = -32; this is the acceleration due ot gravity (again, until the shell hits the ground.) Find y(t).

d. Find the Initial Time (t sub 0) that the shell hits the ground, and the x position x(t sub 0) at that time (This position is called the range).

So far, I think I have figured out C, which is y'(t)= -32t+C, C = 200. Other than that, I am lost! Thanks alot in advance!
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Hi, Faisan,

This is your basic 'vertical motion' problem even though the motion isn't strictly vertical.  As far as the horizontal motion is concerned, it is constant until, as you noted, the shell hits the ground.  So we can determine the 'range' of the shell by figuring out how long it stays in the air, multiplied by the horizontal velocity.

For the time being, ignore the horizontal part.

The shell is fired upwards from the ground (making  y0 = 0) at an initial velocity (v0) of  400 sin(pi/6).

So  v0 = 400 sin(pi/6) =  400(1/2) = 200.

For the vertical motion, the basic equation is:

y = -1/2 g t^2 + v0 t + y0

And here, that gives:

y = -16t^2 + 200t

And yes, y' = -32t + 200  (you got this right.)

Now all we ask is 'when does the shell hit the ground?'  Answer: When  y = 0.

And when, pray tell, is that?

Solve  -16t^2 + 200t = 0, a quadratic.

8t(-2t + 25) = 0

That gives  t = 0  (the initial time, when it was fired.)
and  t = 25/2

That's when it returns (strikes the ground).

Now you need the horizontal velocity,  x' = 400 cos(pi/6) = 200(sqrt(3)/2) = 200 sqrt(3)

OK, that's it.  The range is  t * x' = (25/2) 200 sqrt(3) = 2500 sqrt(3)

You can use your calculator on that.