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
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