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ActivPhysics OnLine Problem 3.3: Changing the x-velocity

ActivPhysics OnLine Problem 3.3: Changing the x-velocity Choose xy coordinates in the usual way. y x upwards : + y downwards: - y In these questions we are assuming there is no air resistance. Problem 3.3: Question 1 What is the ball’s total time in the air?

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ActivPhysics OnLine Problem 3.3: Changing the x-velocity

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  1. ActivPhysics OnLine Problem 3.3: Changing the x-velocity Choose xy coordinates in the usual way. y x upwards : + y downwards: - y In these questions we are assuming there is no air resistance.

  2. Problem 3.3: Question 1 What is the ball’s total time in the air? First let us find the time for the ball’s upward trajectory. Call this t. The ball’s initial velocity in the y direction is +10 m/s, but at the highest point of the ball’s motion its velocity in the y direction is 0.

  3. viy= +10 m/s g = -10 m/s2 v= viy + ay t 0= +10 m/s + (-10 m/s2) t 10 m/s2 t = +10 m/s t = 1.0 s Remember that this is the time for the ball’s upward trajectory. If there is no air resistance, then the ball’s path is a parabola (which is symmetric) and the time for the ball’s downward trajectory is also 1.0 s. Thus the ball’s total time in the air is 2 s.

  4. The ball’s total time in the air has the same value 2.0 seconds and does not depend on the velocity in the x direction because the motions in the x and y directions are independent. The ball’s total time in the air depends only on its initial y position, its initial velocity in the y direction, and its acceleration in the y direction.

  5. Question 2: What is the ball’s maximum height? v2= viy2 + 2 a (yf - yi ) 0 = (+10 m/s)2 +2 (-10 m/s2) (yf -0) 2 (10 m/s2) yf = (10 m/s)2 yf = 5.0 m The ball’s maximum height has the same value 5.0 meters, regardless of its velocity in the x direction, because motions in the x and y directions are independent.

  6. The ball’s maximum height depends only on the ball’s initial y position, its initial velocity in the y direction, and its acceleration in the y direction.

  7. Question 3: What is the horizontal distance or range that the ball travels before hitting the ground? x = vx t + ½ (ax t2) There is no acceleration in the x direction, ax = 0, and so x = vx t. The horizontal range x is greater if the horizontal component of the ball’s velocity is greater. If vx = 2.0 m/s, then x = (2.0 m/s)(2.0 s) = 4.0 m. If vx = 10 m/s, then x = (10 m/s)(2.0 s) = 20 m.

  8. ActivPhysics OnLine Problem 3.4: x- and y- accelerations of a projectile Choose xy coordinates in the usual way. y x Question 1: What is the acceleration in the x direction? Ans.: Zero.

  9. Question 2: Calculate the acceleration y components from the graph. t (s)v (m/s) 0.0 +30 1.0 +20 2.0 +10 3.0 0.0 4.0 -10 a = (vf – vi)/ (tf – ti) For example from 0 to 1 s a = (20 m/s - 30 m/s)/(1s – 0s) a = -10 m/s2 In each case this same value is obtained.

  10. Question 3: By how much does the y-velocity component change in each of the following time intervals? ay = -10 m/s2 vy = viy + ay t Δvy = ay t For t = 1.0 s, Δvy = -10 m/s. For t = 2.0 s, Δvy = -20 m/s. For t = 0.1 s, Δvy = -1.0 m/s.

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