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Vertical motion of a baseball

A batter hits a baseball straight upward at home plate and the ball is caught 5.0 s after it is struck [link] . (a) What is the initial velocity of the ball? (b) What is the maximum height the ball reaches? (c) How long does it take to reach the maximum height? (d) What is the acceleration at the top of its path? (e) What is the velocity of the ball when it is caught? Assume the ball is hit and caught at the same location.

Left picture shows a baseball player hitting the ball at time equal zero seconds. Right picture shows a baseball player catching the ball at time equal five seconds.
A baseball hit straight up is caught by the catcher 5.0 s later.

Strategy

Choose a coordinate system with a positive y -axis that is straight up and with an origin that is at the spot where the ball is hit and caught.

Solution

  1. [link] gives
    y = y 0 + v 0 t 1 2 g t 2

    0 = 0 + v 0 ( 5.0 s ) 1 2 ( 9.8 m / s 2 ) ( 5.0 s ) 2 ,

    which gives v 0 = 24.5 m/sec .
  2. At the maximum height, v = 0 . With v 0 = 24.5 m/s , [link] gives
    v 2 = v 0 2 2 g ( y y 0 )

    0 = ( 24.5 m/s ) 2 2 ( 9.8 m/s 2 ) ( y 0 )

    or
    y = 30.6 m .
  3. To find the time when v = 0 , we use [link] :
    v = v 0 g t

    0 = 24.5 m/s ( 9.8 m/s 2 ) t .

    This gives t = 2.5 s . Since the ball rises for 2.5 s, the time to fall is 2.5 s.
  4. The acceleration is 9.8 m/s 2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s 2 downward.
  5. The velocity at t = 5.0 s can be determined with [link] :
    v = v 0 g t = 24.5 m/s 9.8 m/s 2 ( 5.0 s ) = −24.5 m/s .

Significance

The ball returns with the speed it had when it left. This is a general property of free fall for any initial velocity. We used a single equation to go from throw to catch, and did not have to break the motion into two segments, upward and downward. We are used to thinking of the effect of gravity is to create free fall downward toward Earth. It is important to understand, as illustrated in this example, that objects moving upward away from Earth are also in a state of free fall.

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Check Your Understanding A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantity increases faster, the speed of the ice chunk or its distance traveled?

It takes 2.47 s to hit the water. The quantity distance traveled increases faster.

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Rocket booster

A small rocket with a booster blasts off and heads straight upward. When at a height of 5.0 km and velocity of 200.0 m/s, it releases its booster. (a) What is the maximum height the booster attains? (b) What is the velocity of the booster at a height of 6.0 km? Neglect air resistance.

Figure shows a rocket releasing a booster.
A rocket releases its booster at a given height and velocity. How high and how fast does the booster go?

Strategy

We need to select the coordinate system for the acceleration of gravity, which we take as negative downward. We are given the initial velocity of the booster and its height. We consider the point of release as the origin. We know the velocity is zero at the maximum position within the acceleration interval; thus, the velocity of the booster is zero at its maximum height, so we can use this information as well. From these observations, we use [link] , which gives us the maximum height of the booster. We also use [link] to give the velocity at 6.0 km. The initial velocity of the booster is 200.0 m/s.

Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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