1. ## projectile 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?

2. Originally Posted by Jenny20
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?
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

3. Hello, Jenny!

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