Question

The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 ft./s is given by the equation H equals 3+90 T -16 T squared where t equals time in seconds. After how many seconds will the ball be 84 feet above the ground

Answer 1

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

The height $h(t)$ of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation $h(t) = 3 +90\cdot t -16\cdot t^{2}$, where $t$ is time in seconds. After how many seconds will the ball be 84 feet above the ground.

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

$3+90\cdot t -16\cdot t^{2} = 84$

$16\cdot t^{2}-90\cdot t +81 = 0$ (1)

By Quadratic Formula:

$t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}$

$t_{1} = 4.5\,s$, $t_{2} = 1.125\,s$

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.