29. BASKETBALL When Jerald shoots a free throw, the ball is 6 feet from the floor and has

an initial upward velocity of 20 feet per second. The hoop is 10 feet from the floor.

a. Use the vertical motion model to determine an equation that models Jerald's

free throw.

b. How long is the basketball in the air before it reaches the hoop?

c. Raymond shoots a free throw that is 5 foot 9 inches from the floor with the same

initial upward velocity. Will the ball be in the air more or less time? Explain.

  a) j(t) = -16t^2 +20t +6

  b) 1 second

  c) less time

Step-by-step explanation:

The general equation for vertical motion is ...

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a) j(t) = -16t^2 +20t +6 . . . . . filling in the given initial values, with g=32 ft/s^2

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b) We want to find the larger of the solutions to ...

  j(t) = 10 = -16t^2 +20t +6

Subtracting 6 and dividing by -4, we have ...

  -1 = 4t^2 -5t

  4t^2 -5t +1 = 0 . . . . add 1, put in standard form

  (4t -1)(t -1) = 0 . . . . . factor

  t = 1/4 or 1

The ball is in the air 1 second before reaching the hoop.

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c) In generic terms, the equation of motion for a given shooting height at the same upward velocity will be ...

  10 = -16t^2 +20t + h0

  -16t^2 +20t +(h0 -10) = 0

Using the quadratic formula, we find the larger solution to be ...

  t = (-(20) -√(20² -4(-16)(h0 -10)))/(2(-16))

  t = (5/8) + √(64h0 -240)/32 = (5 +√(4h0 -15))/8

This value increases as h0 increases, so the ball will be in the air less time when Raymond shoots.

you can't solve x because it is 2 linear equation and you only have one equation

22.83 repeating

step-by-step explanation:

137 ÷ 6 is 22.8333333 therefore we understand that 3 is repeating.

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