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A major league baseball pitcher throws a pitch that follows these parametric equations:

x(t) = 143t
y(t) = –16t2 + 5t + 5.
Recall that the speed of the baseball at time t is

s(t)=√ [x '(t)]2 + [y ' (t)]2 ft/sec.

What is the speed of the baseball (in mph) when it passes over homeplate?

Respuesta :

W0lf93
97.67 mph The distance between the pitches plate and homeplate is 60.5 ft. So the time t at which the ball passes over home plate will be 60.5 = 143t 0.423 = t Now calculate the first derivative of each equation. So x(t) = 143t x'(t) = 143 y(t) = -16t^2 + 5t + 5 y'(t) = -32t + 5 So at 0.423 seconds, the respective velocities will be x'(0.423) = 143 y'(0.423) = -32 * 0.423 + 5 = -13.536 + 5 = -8.536 And the speed will be sqrt(143^2 + (-8.536)^2) = sqrt(20449 + 72.8633) = sqrt(20521.86) = 143.2545 ft/s Now convert from ft/s to mile/hour 143.2545 ft/s * 3600 s/hour / 5280 = 97.67 mile/hour = 97.67 mph

The speed of the baseball when it passes over the home plate is 97.67 mph.

Given :

  • [tex]x(t) = 143t[/tex]    ---- (1)
  • [tex]y(t) = -16t^2+5t+5[/tex]    ---- (2)
  • The distance between the pitcher's plate and the home plate is 60.5 ft.

Differentiate the function x(t) and y(t) with respect to t.

[tex]x'(t)=143[/tex]

[tex]y'(t)= -32t+5[/tex]   ---- (3)

Now, put the value of x(t) in equation (1).

60.5 = 143t

t = 0.423 sec

Now, put the value of t in equation (3).

[tex]y'(t) = -32(0.423)+5=-8.536[/tex]

Now, [tex]s(t) = \sqrt{(x'(t))^2+(y'(t))^2}[/tex]

[tex]s(t)=\sqrt{143^2+(-8.536)^2}[/tex]

[tex]\rm s(t)=\sqrt{20521.8633} = 143.2545\;ft/sec[/tex]

[tex]\rm s(t) = 97.67 mph[/tex]

Therefore, the speed of the baseball when it passes over the home plate is 97.67 mph.

For more information, refer to the link given below:

https://brainly.com/question/11897796

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