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Free-Fall Motion As a promotion for the Houston Astros downtown ballpark, a competition is held to see who can throw a baseball the highest from the front row of the upper deck of seats, 83 ft above field level. The winner throws the ball with an initial vertical velocity of 92 ftsec and it lands on the infield grass.
(a) Find the maximum beight of the baseball.
(b) How much time is the ball in the air?
(c) Determine its vertical velocity when it hits the ground.
Baseball Throwing Machine The Sandusky Little League uses a baseball throwing machine to help train 10-year- old players to catch high pop-ups. It throws the baseball straight up with an initial velocity of 48 ft/sec from a height of 3.5 ft.
(a) Find an equation that models the height of the ball I sec- onds after it is thrown.
(b) What is the maximum height the baseball will reach? How many seconds will it take to reach that height?

Respuesta :

1) The respective answers for the maximum height, time in the air and vertical velocity are; 215.25 ft; 6.54 s; -117.28 ft/s

2) The respective answers for the equation that models the height of the ball and the value of maximum height of the ball are;

s(t) = -16t² + 48t + 3.5

Maximum height = 39.5 ft

Quadratic parabola

1) (a) We are given that;

  • Highest front row above field level = 83 ft
  • Initial Vertical Velocity = 92 ft/s

The equation that represents the height at time t is;

S(t) = -16t² + 92t + 83

The maximum height will occur at the time of the x-value of the line of symmetry of the parabola.

Thus; t = -b/(2a) = -92/(2 * -16)

t = 2.875 s

Thus, maximum height is;

S(2.875) = -16(2.875)² + 92(2.875) + 83

S(2.875) = 215.25 ft

1b) The time that the ball is in the air is at S(t) = 0 ft.

Thus;

-16t² + 92t + 83 = 0

From online quadratic equation calculator;

t = 6.54 s

1c) The vertical velocity when it hits the ground is gotten by differentiating the height equation and putting 6.54 s for t. Thus;

v(t) = s'(t) = -32t + 92

v(6.54) = -32(6.54) + 92

v(6.54) = -117.28 ft/s

2a) We are given;

  • Initial velocity = 48 ft/s
  • height = 3.5 ft

Thus, equation that models  the height of the ball t seconds after it is thrown is;

s(t) = -16t² + 48t + 3.5

2b) The maximum height will occur at the time of the x-value of the line of symmetry of the parabola.

Thus; t = -b/(2a) = -48/(2 * -16)

t = 1.5 s

Thus, maximum height is;

S(2.875) = -16(1.55)² + 92(1.5) + 3.5

S(2.875) = 39.5 ft

Read more about quadratic parabola at; https://brainly.com/question/14477557

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