Question

A soccer ball is kicked into the air from a height of 3 feet with an initial velocity of 72 feet per second. How many seconds is the ball 84 feet above the ground?

Answer 1

Answer:

For 2.25 seconds the ball is 84 feet above the ground.

Step-by-step explanation:

Initial height when the ball is kicked = 3 feet

Initial velocity of the soccer ball = 72 feet per second

The height(h) of the vertical thrown object is modeled by the formula :

h(t) = -16t² + vt + s , where t is the time, v is initial velocity and s is initial height

Now, h(t) is given to be 84 feet and we need to find the time taken by the ball to reach 84 feet above the ground :

84 = -16t² + 72t + 3

⇒ 16t² - 72t + 81 = 0

Finding roots of this equation :

$t = \frac{72\pm\sqrt{72^2-4\times 16\times 81}}{2\times 16}\\\\\implies t = \frac{72\pm 0}{32}\\\\\implies t = 2.25$

Hence, for 2.25 seconds the ball is 84 feet above the ground.