Phy 231

A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.

What will be the velocity of the ball after one second?

answer/question/discussion: vf = v0 + a * 'dt

vf = 25 m/s + (-10 m/s^2 * 1 s) = 25 m/s - 10 m/s = 15 m/s

What will be its velocity at the end of two seconds?

answer/question/discussion: vf = v0 + a * 'dt

vf = 25 m/s + (-10 m/s^2 * 2 s) = 25 m/s - 20 m/s = 5 m/s

During the first two seconds, what therefore is its average velocity?

answer/question/discussion: vAve = (vf + v0) / 2

vAve = (5 m/s + 25 m/s) / 2 = 30 m/s / 2 = 15 m/s

How far does it therefore rise in the first two seconds?

answer/question/discussion: 'ds = vAve * 'dt

'ds = 15 m/s * 2 s = 30 m

What will be its velocity at the end of a additional second, and at the end of one more additional second?

answer/question/discussion:

After 3 seconds:

vf = v0 + a * 'dt

vf = 25 m/s + (-10 m/s^2 * 3 s) = 25 m/s - 30 m/s = -5 m/s

After 4 seconds:

vf = v0 + a * 'dt

vf = 25 m/s + (-10 m/s^2 * 4 s) = 25 m/s - 40 m/s = -15 m/s

At what instant does the ball reach its maximum height, and how high has it risen by that instant?

answer/question/discussion: The ball should be at its max height when vf = 0 m/s.

a = (vf - v0) * 'dt, or 'dt = (vf - v0) / a

'dt = (0 m/s - 25 m/s) / -10 m/s^2

'dt = -25 m/s / -10 m/s^2 = 2.5 s

Now that we know 'dt = 2.5 s, we can find 'ds:

(vf + v0) / 2 * 'dt = 'ds

'ds = (0 m/s + 25 m/s) / 2 * 2.5 s = 31.25 m

What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?

answer/question/discussion: After 4 s, vf = -15 m/s, therefore:

vAve = (vf + v0) / 2 = (-15 m/s + 25 m/s) / 2 = 5 m/s

'ds = vAve * 'dt = 5 m/s * 4s = 20 m

How high will it be at the end of the sixth second?

answer/question/discussion: After 6 s, vf = -35 m/s, therefore:

vAve = (vf + v0) / 2 = (-35 m/s + 25 m/s) / 2 = -5 m/s

'ds = vAve * 'dt = -5 m/s * 6s = -30 m

25 mins