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Stan the Man, from the local circus, was a human cannon ball. While practicing one day, he was shot out of a cannon and intoReally need help with 1) How fast was Stan going at the time of impact?

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Answer #1

To find the angle at which Stan was projected:

We are given the length of the cannon and also how high off the ground one end was. we can use that along with basic trigonometry to fond the projectile angle

so we have sin \theta = \frac{4}{5}

therefore in5 53.1353°

and we are also given the initial velocity u = 25 m/s

For a projectile motion from a cliff we shall refer to this general schematic:

So R = 200 m

H = 250 m

v = 25 m/s we can resolve this into two components ul u- u cos 0 Resolution of components of velocity ()

  7L

and   ITL

and \theta = 53 \degree

We'll use another kinematics equation to determine the final y-velocity.

v^{2}_{f,y} = v^2_{y} + 2 \ g \ H \ \ \ \ \ \ \ ...(1) where g = 9.8 m/s2 is the acceleration in the y- direction (due to gravity)

therefore 1,2,, 202 2 × 9.8 × 250

so U5300

therefore ツ,,-V5300 = 72.8

The x-velocity can also be determined using a similar equation to (1)

here the x-direction acceleration is not given (usually we have to consider the drag due to wind but we are going to ignore that in this problem) so since there is no acceleration is x direction the x component of velocity stays the same.

uf,,-u! = 15 m

Now we shall find the magnitude of the final velocity using the Pythagorean theorem:

TL vf- V152+72.82 74.32

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