ActivPhysics OnLine Problem 3.3: Changing

the x-velocityChoose xy coordinates in

the usual way. y

x upwards : + y downwards: - yIn 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?

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.

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.

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.

Question 2: What is the ball’s maximum height?

v2 = viy 2 + 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 mThe 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.

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.

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.

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.

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.

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.