We will model the flight path of a baseball by the parametric equations, X (t) =

Question

We will model the flight path of a baseball by the parametric equations,

X(t)=V0cos(€)t-12t^2

Y(t)=-16t^2+V0sin(€)t+3-10t

Where the additional terms have been included to stimulate the time in flight, the influence of the wind, and the height of the ball at impact (3 feet). This model is still not quite physically correct, for resistance to wind is often regarded as proportional to velocity raised to some power yielding a different and more complex model.

2.) With bases loaded, a full count, and two cuts in the bottom of the ninth, Joe’s team is down two runs. Joe makes contact with the next pitch giving it 116mph initial velocity and a 250 angle of elevation. The ball heads directly towards center field where the 10 feet tall fence is 404 ft from home plate. Did Joe hit a home run?

Explanation / Answer

X = Vo*T*COS(E)-12*T^2 Y= - 16*T^2 +Vo*T*SIN(E)+3-10*T Vo= 116 MPH Vo= 170.1333 FPS E=250 DEG.= 2.181662 RADIANS PSE CHECK UP FOR TYPO? ANGLE OF ELEVATION SHALL BE BELOW 90 DEGREES & AROUND 45 FOR MAXIMUM LIFT . FENCE COORDINATES ARE …. Xf= 404 FT Yf= 10 FT. LET US FIND T & WHEN IT REACHES THE FENCE … BY TABULATING THE RESULTS AS FOLLOWS…. T X Y 0 0 3 0.1 -9.87845 15.77651 SO WITH THE DATA GIVEN THE BALL NEVER REACHES THE FENCE .. TAKING THE ANGLE AS 50 DEGREES , WE GET …. E=50 DEG. = 0.436332 RADIANS WE GET THE NEW TABULATION AS T X Y 0 0 3 1 142.1932 48.90145 2 260.3863 62.80291 3 354.5795 44.70436 3.66603815 404 14.89572 SO WHEN THE BALL REACHES THE FENCE AT X=404' FROM HOME PLATE , ITS ELEVATION Y=14.895 ' > 10 ' NEEDED TO MAKE A HOME RUN SO JOY HITS THE HOME RUN IF THE ANGLE OF ELEVATION IS 50 DEG. & NOT REPEAT NOT 250 DEG. WHICH IN ANY CASE IS PATENTLY ABSURD.

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