Calculating problems with x and y components and other questions

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SUMMARY

The discussion focuses on calculating the kinetic energy and velocity components of projectiles using equations of motion. The equations provided include x(t) = a + b*t and y(t) = c + d*t + e*t*t, with specific values for a, b, c, d, and e. Participants discuss the application of the Pythagorean theorem for determining resultant velocities and kinetic energy, emphasizing the importance of understanding the correct application of forces and acceleration in projectile motion. Key corrections include recognizing that -9.8 m/s² is acceleration, not force, and ensuring proper use of the kinetic energy formula KE = 0.5 * m * v².

PREREQUISITES
  • Understanding of projectile motion equations
  • Familiarity with kinetic energy calculations
  • Knowledge of vector components and the Pythagorean theorem
  • Basic principles of forces and acceleration in physics
NEXT STEPS
  • Study the derivation and application of the equations of motion for projectiles
  • Learn about the Pythagorean theorem in the context of vector addition
  • Review the principles of kinetic energy and its calculation in physics
  • Examine the effects of gravitational acceleration on projectile motion
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Students and educators in physics, engineers working with projectile dynamics, and anyone interested in understanding the calculations involved in kinetic energy and motion analysis.

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1.)

x(t) = a + b*t

y(t) = c + d*t + e*t*t

a= 70 cm, b=38 cm/s, c=78 cm, d=17 cm/s and e=-490 cm/s/s

What is the initial velocity of this projectile?


I approached it assuming that the angle is 45 degrees. So I used the pythagorean therom to solve for the hypotnuse given that we already have the vertical and horizontal velocities. Is this correct or am I missing something here?

2.) A 510 gram ball is moving through the air. At this moment, the horizontal component of its velocity is vx=9.43 m/s and the vertical component of its velocity is vy=4.70 m/s.

What is its kinetic energy now?


I did an online quiz and used the pythagorean theorem again. Somehow I got the right answer on another exact problem with different numbers. Is the pythagorean only involved in this problem?

3.) A bullet is being shot straight up with a mass of 35.58gm and a velocity of v=176m/s. What is its kinetic energy in 47.91 seconds?

For this one, I assumed a force of -9.8m/s/s was present on the bullet being shot straight up. I multiplied 47.91 to -9.8, and added that value to 176. The outcome of this should be the instant where we need to find the KE. I than used the KE=.5M+(V*V) formula to find the answer. Am this the correct method to solving it? The online quiz keeps telling me I've been getting the wrong answer.
 
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1) Out of the 5 constants listed (a-e), which have the units of velocity?
Remember one way to write a position equation is
[itex]x(t) = x_0 + v_0t + \frac{1}{2} gt^2[/itex], what do you notice about each component and the power of t? Once you've discovered that, find the numerical value for each of the velocity-variables, which are already given in components. Combine the two vector components to obtain the vector's direction and its magnitude.

What is its kinetic energy now?
You can find an expression for kinetic energy in your book. You are given the velocity in two components again, which you'll combine to find the magnitude of the velocity vector (you'll need it in your KE expression). You always need pythag theorem to combine components.

3.) A bullet is being shot straight up with a mass of 35.58gm and a velocity of v=176m/s. What is its kinetic energy in 47.91 seconds?
To find the kinetic energy, you'll need the object's instantaneous velocity at t=47.91 sec. First off,
I assumed a force of -9.8m/s/s was present
-9.8m/s^2 is an acceleration not a force.

Your expression for KE is incorrect, close though. Check it again in your book.
 

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