A shell leaves a mortar with a muzzle velocity of 500 ft/s directed upward at 60 degrees with the horizontal. Determine the position of the shell and its resultant velocity 20 s after firing. how high will it rise?
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Depends,
If there gravity (i'll assume there is) then the shell's upward velocity changes over time.
If there is air drag (I'll assume there is NOT) then the shell's velocity in all axes changes over time.
So, no drag, but gravity.
First we break the shell's starting velocity (Vo) into two vectors, upward, and downrange, or y and x. We will call them Initial Velocity Vertical (Vov) and Inital Velocity Horizontal or Voh.
(Most sub and superscripts don't work here, this is easier...)
Since Sin Θ = Opposite / Hypotenuse and Cos Θ = Adjacent over Hypotenuse
Sin Θ = Vov / Vo and Cos Θ = Voh / Vo
Sin 60 = Vov / 500 and Cos 60 = Voh / Vo
sin 60 * 500 = Vov and .5 * 500 = Voh
Therefore vertical starting velocity Vov = 433.0127 ft/sec
And horizontal starting velocity Voh = 250 ft/sec
To check Vov^2 + Voh^2 should = Vo^2 (pythagorean theorum)
433.0127^2 + 250^ = 249,999.99836129
Which is close enough for government work.
If you want more accurate answers, instead of calculating Vov, use the formula that generates it in it's place. (which I did below)
Horizontal Position.
Velocity * Time = Distance.
Since horizontal velocity doesn't change (no drag), 20 seconds after firing, the shell is 250 * 20 feet downrange.
250*20 = 5,000 feet.
Vertical position.
Now because there is an acceleration being applied to the shell, (gravity at -32 ft/sec^2 (assumed approximate value, should be stated) ) the formula for position vertically is:
y = 1/2 a * t^2 + (Vov * t)
So we just plug and play
y = 1/2 * -32 * 20^2 + ((sin 60 * 500 ) * 20) = 2,260.25404 ft above firing point.
so the ordered pair for the location of the shell @ 20 sec after firing is
(5000ft, 2,260.25404 ft)
If you're looking for something like "displacement from firing position" you're going to have to use the pythagorean theorum to find it...
(BTW: flight time on this shell in this trajectory is ~27 seconds).
If there gravity (i'll assume there is) then the shell's upward velocity changes over time.
If there is air drag (I'll assume there is NOT) then the shell's velocity in all axes changes over time.
So, no drag, but gravity.
First we break the shell's starting velocity (Vo) into two vectors, upward, and downrange, or y and x. We will call them Initial Velocity Vertical (Vov) and Inital Velocity Horizontal or Voh.
(Most sub and superscripts don't work here, this is easier...)
Since Sin Θ = Opposite / Hypotenuse and Cos Θ = Adjacent over Hypotenuse
Sin Θ = Vov / Vo and Cos Θ = Voh / Vo
Sin 60 = Vov / 500 and Cos 60 = Voh / Vo
sin 60 * 500 = Vov and .5 * 500 = Voh
Therefore vertical starting velocity Vov = 433.0127 ft/sec
And horizontal starting velocity Voh = 250 ft/sec
To check Vov^2 + Voh^2 should = Vo^2 (pythagorean theorum)
433.0127^2 + 250^ = 249,999.99836129
Which is close enough for government work.
If you want more accurate answers, instead of calculating Vov, use the formula that generates it in it's place. (which I did below)
Horizontal Position.
Velocity * Time = Distance.
Since horizontal velocity doesn't change (no drag), 20 seconds after firing, the shell is 250 * 20 feet downrange.
250*20 = 5,000 feet.
Vertical position.
Now because there is an acceleration being applied to the shell, (gravity at -32 ft/sec^2 (assumed approximate value, should be stated) ) the formula for position vertically is:
y = 1/2 a * t^2 + (Vov * t)
So we just plug and play
y = 1/2 * -32 * 20^2 + ((sin 60 * 500 ) * 20) = 2,260.25404 ft above firing point.
so the ordered pair for the location of the shell @ 20 sec after firing is
(5000ft, 2,260.25404 ft)
If you're looking for something like "displacement from firing position" you're going to have to use the pythagorean theorum to find it...
(BTW: flight time on this shell in this trajectory is ~27 seconds).