Question

Jon is hitting baseballs. When he tosses the ball into the air, his hand is 5 feet above the ground. He hits the ball when it falls back to a height of 4 feet. The height of the ball is given by h = - 16x^2 + 6t + 5 where in seconds How much time will pass before Jon hits the ball? What is the maximum height the ball attains?

Answer 1

Answer:

Time  that will pass before Jon hits the ball is 0.5 seconds  

The maximum height the ball attains is 6.875 feet

Step-by-step explanation:

Step 1: To find out how much time passes before Jon hits the ball

The height he hits the ball when it falls is 4 feet

so,

$4 = - 16t^2 + 6t + 5$

$4 + 16t^2 - 6t - 5 = 0$

$16t^2 - 6t -1 = 0$ -------------------------(1)

Solving eq(1) using quadratic formula, we get

$t = \frac{-b \pm\sqrt{b^2-4ac}}{2a}$

$t = \frac{-(-6) \pm\sqrt{(-6)^2-4(16)(-1)}}{2(16)}$

$t = \frac{6 \pm\sqrt{36+64}}{32}$

$t = \frac{6 \pm\sqrt{100}}{32}$

$t = \frac{6 \pm10}{32}$

Taking only the positive value

$t = \frac{16}{32}$

t = 0.5 seconds

Step 2: To find the maximum height of the ball

Max height  will be reached at [$\frac{-b}{2a}$] sec

$=\frac{-b}{2a}$

$\\ = \frac{-(-6)}{2\times16}$

$= \frac{6}{32}$

Now the height at t= $\frac{6}{32}$ is

$16( \frac{6}{32})^2 - 6( \frac{6}{32}) -1 = h$

$( \frac{6}{2})^2 - \frac{36}{32} -1 = h$

$( 3)^2 - 1.125 -1 = h$

$9 - 1.125 -1 = h$

6.875  =  h