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There are two ways in which this problem can be solved. First way is to write the equation of motion for an object that is thrown up with initial speed `v_0` subjected to gravitational acceleration `a =32.17 ft/s^2` .
`h =h_0+ v_0*t-(32.17/2)*t^2`
and by inserting the values for time and height given in the text one can determine a medium initial speed `v_0` and `h_0` from the two by two systems of equations that form.
A second more elegant way is to graph the points data given in text and to fit them with a quadratic curve
`f(x) = A +B_1*X +B_2*X^2`
which correspond to the general equation of motion
`H =H_0 +V_0*t +(g*t^2)/2`
The curve and fit are presented in the figure attached below. The coeffcients of the curve that fits the given points are
`A =H_0 =3 ft`
`B_1 =V_0 =110.5 ft/s`
`B_2 =-g/2 =-15.9 ft/s^2` which gives a value `g =-31.8 ft/s^2` .
Thus the equation of the motion is
`H =3+110.5*t -(31.8/2)*t^2`
which gives for the time to maximum height the value
`t_("max") =V_0/g = 110.5/31.8 =3.475 s`
and a maximum height of
`H_max =3+110.5*3.475 -(31.8/2)*3.475^2 =194.98=195 ft`
Answer: initial speed is `V_0 =110.5 ft/s` and height of the ball when hit is `H_0 =3 ft ` .
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