An object is thrown straight up at 100 m/s. how fast is it traveling 3 seconds later? how long after throwing?

Question

an object is thrown straight up at 100 m/s. how fast is it traveling 3 seconds later? how long after throwing it up in the air will it reach its maximum height, stop moving and then be about to start to fall? show your work by using a drawing and/ or calculations to demonstrate your answer/explanation

We are using 10 instead of 9.8

Answer 1

Best Answer - Chosen by Asker

V = vi - g*t = 100 - 10*3 = 70 m/s

t = (vi - V)/g = vi/g = 10 sec

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Asker's Comment:

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Answer 2

It depends on the size of the ball because of air resistance: how much friction is exerted on the ball by the molecules of the air it's passing through. The questioner said it was thrown up "in the air" so, without stating the size of the ball or other relevant variables like the altitude above sea level from which the ball was launched (air density varies with altitude so, so would the air resistance), it is actually not possible to say for sure. Your questioner only specified enough variables to make the necessary calculations if you assume the ball is projected vertically in a vacuum. But they specified "in the air". I would question the question but it wouldn't make your questioner a very 'happy bunny' so better just going with the vacuum assumption and saying 70 and 10 to get the marks before deciding whether to tell the questioner they're not very good at specifying the necessary parameters within which they intend to judge your ability.

Answer 3

In "Free-Fall" any object will vertically lose 9.8 m/s of upward speed every second.
In 3 seconds, the object loses upward speed = (9.8)(3) = 29.4 m/s,
so it is moving at 100 - 29.4 = 70.6 m/s upward after 3 seconds ANS-1

Object reaches max height when its upward speed = 0, at 100/9.8 = 10.2 s ANS-2