Determine the initial velocity of the baseball and the height of the ball when hit.Time(sec) .75 1.5 2 2.75 3.25 4.75 Height(ft) 77 133 160 187 194 169
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 ` .