Question

A baseball leaves the hand of a pitcher 6 vertical feet above home plate and 60 feet from home plate. Assume the x-axis is a straight on the flat ground from mound to the plate, the y-axis lies on the groundfrom the 3rd to 1st base and the z axis is vertical height.
(a) In the absence of all forces except gravity, assume that a pitch is thrown with an initial velocity of 130i +0j-3k ft/s (roughly 90mi/hr). How far above the ground is the ball when it cross the home plate.
(b) How long does it take for the pitch to arrive at home plate?
(c) what vertical velocity component should the pitcher use so that the pitch crosses home plate exactly 3 feet above the ground?
(d) A simple model to describe the curve of the baseball assumes that the spin of the ball produces a constant acceleration (in the y direction) of c ft/s^(2). Assume a pitcher throws a curve ball with c=8ft/s^(2) (1/4 the acceleration of gravity). how far does the ball move inthe y-directionby the time it reaches home plate, assuming an initial velocity of 130i +0j -3k ft/s ?
(e) Using the above information, does the ball curve more in the first half of its trip to second or in the second half?
(f) How does this effect better?
(g) Suppose the pitcher releases the ball from an initial position of <0,-3,6> with an initial velocity of 130i +0j-3k. What value of the spin parameter c is needed to put the ball over home plater passing through the point <60,0,3>?

Answer 1

Step-by-step explanation:

Throughout the solution we will use seconds for time, feet for distance, ft/sec for speed, and ft/sec2  for  acceleration.

We are assuming that r(0) = (0, 0, 6), v(0) = (130, 0, −3), and a(t) = (0, 0, −32).

Note: These are all vectors in i,j and k directions.

We need to work from a(t),  to v(t), to r(t) by integration.

Integrate a(t) to find v(t)

$v(t) = \int\limits {a(t) } \, dt = \int\limits {(0,0,-32)} \, dt = (0,0,-32) *t + C$

Using v(0) = (130, 0, −3), we can determine C

C = (130, 0, −3)

Hence, v(t) = (130, 0, 32*t -3) ...... Eq1

Integrate v(t) to find r(t)

$r(t) = \int\limits {(130,0,-3-32*t)} \, dt = (130t , 0 , -3t -16t^2) + C$

Using r(0) = (0, 0, 6), we can determine C

C = (0, 0, 6)

Hence, r(t) = (130t, 0, -3t -16t^2 +6)  ...... Eq2

Solution

The ball crosses home plate when the x-position is 60 feet using Eq 2

130 t = 60

t = 0.462 seconds

@ t = 0.462 the z-coordinate will represent the height of the ball above the ground

$-3 (0.462) -16 (0.462)^2 +6=1.207 ft$

If we rather wanted this height to be 3 feet when the ball crosses the plate then the vertical component  of v(0) will need to be something other that −3.

If we repeat the calculation of r(t) above with v(0) = (130,0,k) then we will find that r(t) = (130t , 0 , 6 + kt - 16t^2)

The ball will still cross home plate at t = 6/13 sec and the  height of the ball as it crosses the plate will be :

$k (0.462) -16 (0.462)^2 +6=3 ft\\\\k = \frac{23}{26}$

d) In this case we are assuming that a(t) = (0, 8, −32). If we repeat the calculation of r(t) above with  a(t) = (0, 8, −32) and v(0) = v(0) = (130, 0, −3), the only change will be in the y-coordinate.

r(t) = (130t, 4t^2, 6 - 3t - 16t^2)

Since the time of flight is still t = 6/13 sec, we calculate that, in the y-direction, the ball has moved (from  t = 0 to t = 6/13):

$y = 4t^2 = 4 (\frac{6}{13})^2 = 0.852 ft$

e) Comparing the first half of the flight (0 ≤ t ≤ 3/13) to the second half of the flight (3/13 ≤ t ≤ 6/13),  the movement in the y-direction is:

$4t^2 = 4 (\frac{3}{13})^2 = 0.213 ft\\4t^2_{1} - 4t^2_{2} = 4 (\frac{6}{13})^2 - 4 (\frac{3}{13})^2= 0.639 ft$

The ball moves more in the second half of its flight.

f) The batter has to adjust to the ball moving sideways  almost eight inches in the last quarter-second of its flight.

g) We will repeat the calculation of r(t), this time with r(0) = (0, −3, 6), v(0) = (130, 0, −3), and a(t) =(0, c, −32). Again, the fact that we are working separately in each coordinate, makes the calculations reasonable.

You can check, that according to the conditions given above

r(t) = (130t, −3 +  0.5ct^2 , 6 - 3t - 16t^2)

At the end of the flight (t = 6/13), the y-coordinate should be 0, so

$-3 + \frac{1}{2}*c*(\frac{6}{13})^2 = 0\\\\c = 28.2 ft/sec^2$