Projectile question please someone help? - cannon diagram, labeled
A projectile from a cannon to 52 degrees above the horizontal at an initial speed of 11m / s and the acceleration of gravity is raised 9.8 m / s 2 ^ The trajectory of the shell curve downward due to gravity at t = 0.383, the tank is below the line by a vertical distance. How do I calculate the distance? In the diagram, there is a firearm during a line at 52 degrees and the maximum height and then start falling a bit on the shells, and I need to get the amount of vertical space, but I have no idea how to create.
Monday, February 8, 2010
Cannon Diagram, Labeled Projectile Question Please Someone Help?
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2 comments:
1) Look how high the tank is at this time.
Consider the vertical direction
You know: V0, A, T
You want, and
Usage: y = v0 * t + ½ A ² t • •
y = (11 sin (52)) (0.383) + ½ (-9.8) (0.383) ² = 2.6 m
See 2) Here, the horizontal distance the shell has traveled:
V = x / tx = ─ ─ ► T • V
x = (11) cos (52) · (0.383) = 2.59 m
3) At an angle of 52 ° to the right to find out what is the value of Y when the value of x 2.59 m:
tan (Θ) = Y / X ─ ─ ► y = x tan (Θ)
y = (2.59) is (52) = 3.32 m
4) Determine the difference between this value and the value found in part (1):
DELTA y = y3 - y1 = 3.32 to 2.6 m = 0.72 = 72 cm
First calculate the first vertical velocity, and then use the equation x = TV + .5 * A * T ^ 2
where x is the vertical distance from the floor compared to the starting point.
v is the first vertical velocity: sin (52) = V/11, V = 11 * sin (52)
a is the acceleration: -9.8 m / s ^ 2 (negative because it is in the opposite direction to the original speed.)
t is the time: 0383
x = (11 * sin (52)) * (0.383) + (0.5) (-9.8) (0.383 ^ 2)
x = 2.60 m
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