A shell, fired from a cannon, has a muzzle speed of 800ft/s. The barrel makes an angle of 45 degree with the horizontal and the barrel opening is assumed to be at ground level.

How far does the shell travel horizontally?

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- Feb 14th 2007, 09:50 AMJenny20projectile motion
A shell, fired from a cannon, has a muzzle speed of 800ft/s. The barrel makes an angle of 45 degree with the horizontal and the barrel opening is assumed to be at ground level.

How far does the shell travel horizontally? - Feb 14th 2007, 10:40 AMtopsquark
I always draw a picture.

So we've got an origin where the shell leaves the cannon (I'll assume it's on the ground at a height of 0 ft.) I am defining +x to the right and +y upward. The initial velocity of the shell is 800 ft/s, at an angle of 45 degrees above the horizontal (to the right in my diagram.)

Note that the shell has no accleration component in the x direction. Thus we know that:

x = x0 + v0x*t

Now, x0 = 0 ft, because the shell started at the origin. v0x = v0*cos(45) = 565.685 ft/s, so

x = 565.685*t

We don't know t.

So let's talk about the motion in the y direction.

y = y0 + v0y*t + (1/2)ay*t^2

Again, y0 = 0 ft, because the shell started at the origin. v0y = v0*sin(45) = 565.685 ft/s, and ay = -32 ft/s^2. So:

y = 565.685*t - 16*t^2

We want the point where the shell hits the ground, so we know that y = 0:

0 = 565.685*t - 16*t^2

Now solve for t:

0 = t(565.685 - 16*t)

Thus t = 0 s or 565.685 - 16*t = 0. The t = 0 s solution is where the shell started, we don't care about this. Thus:

565.685 - 16*t = 0

t = (565.685/16) = 35.355 s.

Going back to the x equation:

x = 565.685*t = 565.685*35.355 = 20000 ft.

-Dan - Feb 14th 2007, 11:34 AMSoroban
Hello, Jenny!

Quote:

A shell, fired from a cannon, has a muzzle speed of 800ft/s.

The barrel makes an angle of 45 degree with the horizontal;

the barrel opening is at ground level.

How far does the shell travel horizontally?

You may be expected to know the "projectile equations".

. . x .= .v(cos θ)t

. . y .= .h + v(sin θ) - 16tē

where: .v = initial speed, h = initial height, θ = angle of elevation.

We are given: .v = 800, h = 0, T = 45°

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._

The equations are: . x .= .(800cos45°)t .= . 400√2 t

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _

. . . . . . . . . . . . . . .y .= .(800sin45°)t - 16tē .= .400√2 t - 15tē

The shell stops when it strikes the ground: .y = 0

. . . . . ._ . . . . . . . . . . . . . . . . . . . . . _

. . 400√2 t - 16tē .= .0 . . → . . 16t(25√2 - t) .= .0

. . . . . . . . . . . . . . _

Hence: .t .= .0, 25√2 seconds

. . . . . . . . . . _ . . . . . . . . . _ . . . ._

When t = 25√2: .x .= .(400√2)(25√2) .= .2000 ft