position function

Question

A varsity soccer player kicks a soccer ball approximately straight upward with an initial velocity of 60 ft/sec. The ball leaves the players foot at a height of 2 feet. Use the position function to write a model representing the height of the soccer ball (in feet) t seconds after being kicked. (Assume the acceleration due to gravity is 32 ft/sec^2) Then use that model to determine the time(s) at which the ball is 52 feet in the air. Use FACTORING to solve. Be sure to write a sentence that explains your answer.

Answer 1

The model representing height is

 

h(t) = (1/2)gt² + v_it + y_i

 

where:

h(t) = the height of the ball

g = acceleration due to gravity ---> this value is negative.

v_i = initial velocity

y_i = initial height

t = time

 

 

h(t) = -16t² + 60t + 2

 

 

To find when the ball is 52 feet in the air, set h(t)=52 and solve for t.

 

52 = -16t² + 60t + 2

 

 

Subtract 52 on both sides of the equation to get a quadratic equation.

 

0 = -16t² + 60t - 50

 

 

Factor out a 2 on the right side.

 

0 = -2(8t² - 30t + 25)

 

 

Factor completely.

 

0 = -2(4t - 5)(2t - 5)

 

 

Set the factors equal to zero.

 

4t - 5 = 0            and              2t - 5 = 0

 

 

Solve for t from the equations.

 

t = 5/4                 and               t = 5/2

 

t = 1.25                and               t = 2.5

 

 

This means that it takes 1.25 seconds to reach a height of 52 feet from the ground and 2.5 seconds to be at that height once more when returning to the ground.