A ball is thrown up into the air at an initial velocity of 50.0 m/s. What is its velocity after 2.50 seconds?

50.0m/s - (2.5s)(9.8m/s

^{2}) =**25.5m/s**-
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50.0m/s - (2.5s)(9.8m/s^{2}) = **25.5m/s**

Lets start by assuming the ball is thrown straight up in the air.

As the ball is thrown up in the air gravity starts to slow it down until the ball changes
**direction** and starts going down and it picks up speed as well. The ball's velocity is constantly changing. Slowing down when going against gravity and speeding up when going with gravity. You must also keep in mind that velocity describes speed
**and** direction. You may notice that after 2.5 seconds the ball is still traveling up. But at what time does it actually change direction?

well the velocity would be zero when it changes direction so just use the equation above. Set velocity to zero and solve for time.

50m/s - t(9.8m/s^{2}) = 0 m/s

50m/s = t(9.8m/s^{2})

50m/s / 9.8m/s^{2} = t

5.01s = t So after 5.01 seconds the ball has come to a complete stop and will soon begin to head down.

Now what is the velocity of the ball after 5.01s?

Lets calculate the velocity at 10.2s. *Why 10.2? You'll see......*

50m/s - (10.2s)(9.8m/s2) = velocity

50m/s - 99.96m/s = velocity

This is a very crucial test question that can go wrong because of a very simple mistake.

Speed and velocity are different things. Keep that in mind.

Good luck!

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