Question

A baseball is thrown up in the air. The table shows the heights y (in feet) of the baseball after x
seconds. Write an equation for the path of the baseball. Find the height of the baseball after 1.7 seconds.

Answer 1

The height of the baseball after 1.7 seconds is 18.85feet

Given the function for calculating the height of the baseball from the table expressed as:

$y=\frac{1}{2}x + 18$ where:

• y is the height of the ball in feet

• x is the time taken in seconds

In order to get the height of the baseball after 1.7 seconds, we will simply substitute x = 1.6 into the expression as shown:

$y=\frac{1}{2}(1.7)+ 18\\y=0.85 + 18\\y= 18.85 feet$

Hence the height of the baseball after 1.7 seconds is 18.85feet

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Answer 2

To answer that question we need to apply the equation for vertical shooting. This particular case implies in fact to steps

• First (going up ) the movement is uniformly decelerated by means of gravity
• Second (going down) the movement is uniformly accelerated again by means of gravity
• g = 9.8 m/s² = 32.17 f/s²

Solution is:

h = 40.085× t -  (1/2)× g× t²  equation for the path

h(₁,₇) = 21.66 f

The annexed table gives information to determine Vo ( initial speed of the ball)

h the path of the ball is:

h = V₀× t - (1/2)×g×t²        

Let´s take the second point of the table  ( 1 ; 24 )

Then  

h = 24   = V₀× (1) - (32.17)×(1)²/2

24 = V₀ - 16.085

V₀  = 40.085  ft/sec

With the value of V₀ we can get the curve (model of the movement)

h = 40.085× t -  (1/2)× g× t²    (1)

Now for t = 1.7 seconds, by substitution en equation (1)

h = 40.085× 1.7 - (1/2)× 32.17× (1.7)²

h = 21.66 f  ( which means the ball is already going down

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