anonymous
  • anonymous
If a ball is thrown upward with an initial velocity of 9.8 m/s, what would be the acceleration, velocity, and position of the ball after a second? 2? 3? (Im interested in the equations and functions and everything..
Mathematics
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katieb
  • katieb
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Math2400
  • Math2400
do u know what kinematics equations are? Or are you supposed to use derivatives? There's a math and physics way but all in all they're similar
anonymous
  • anonymous
Assuming this is on Earth, the acceleration is constant at -9.8 m/s. The Velocity can be calculated using: \[V_{f} = V_{i} + A * T\] Where Vf is the final velocity, Vi is the initial, A is the acceleration, and T is the time. So after 1 second the velocity is 0 m/s After 2 seconds the velocity is -9.8 m/s The Position of the ball (or rather it's distance from the initial point) can be found using: \[D = D_{i} + (V_{i} * T) + (0.5 * A * T^2)\]
anonymous
  • anonymous
I dont know what kinematics are, im taking normal high school physics and i think this came up by accident in my class, but i want to know how to do it, im in calculus and im think i fully understand derivatives

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anonymous
  • anonymous
and where do those equations come fRom?
anonymous
  • anonymous
My high school physics class :) Kinematics, the equations of motion.
anonymous
  • anonymous
I think i know where the first one comes from, but the second one is much more complex, I want to find this using common sense
anonymous
  • anonymous
I thought that the velocity after a second would be 0, but after that is it a constant acceleration, or is this like a parabola?
anonymous
  • anonymous
Think of the acceleration as a rope tied to a sled, the rope is pulling the sled in one direction constantly, in this case, down to the ground. But the velocity is the direction the sled is traveling, and how fast. If an object is traveling upwards at 100 m/s, and the acceleration is downwards at 9.8 m/s, every second that the object is traveling, its velocity decreases by 9.8 m/s. When the velocity reaches 0, the object is motionless, at the top of its path. Every second past that point, the velocity increases downwards by 9.8 m/s until it hits the ground. Acceleration is constant at 9.8 m/s, velocity is modified by the acceleration.
anonymous
  • anonymous
so, after 3 seconds is it 9.8 *2 meters away? or 9.8 squared
anonymous
  • anonymous
You can plug what you know into the distance equation: \[D=D_{i}+(V_{i}∗T)+(0.5∗A∗T^2)\] So: \[D = 0 + (9.8 * 3) + (0.5 * -9.8 * 3^2) Simplifies to: D = -14.7 This means that it is past the starting point, if the starting point was the ground, then it would have hit the ground, otherwise, it would have simply gone backwards past the initial position it was thrown from 14.7 meters.
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