Question

A ball is thrown from a height of 79 meters with initial downward velocity of 6 m/s. The ball’s height in meters after t seconds is given by the following.H=79-6t-5t^2How long after the ball is thrown does it hit the ground

Answer 1

Given the following equation that represents the height of the ball after being thrown.

$\text{ H = 79 - 6t - 5t}^2$

Let's determine the time it'll hit the ground using the quadratic formula:

$\text{ H = 79 - 6t - 5t}^2$

At H = 0,

$\begin{gathered} \text{79 - 6t - 5t}^2\text{ = 0} \\ \text{-5t}^2\text{ - 6t + 79 = 0} \end{gathered}$

a = -5, b = -6 and c = 79

$\text{ t = x = }\frac{-b\text{ }\pm\text{ }\sqrt[]{b^2-4ac}}{2a}$$\text{ = }\frac{-(-6)\text{ }\pm\text{ }\sqrt[]{(-6)^2-4(-5)(79)}}{2(-5)}$$\text{ = }\frac{\text{ 6 }\pm\text{ }\sqrt[]{36\text{ + }1580}}{-10}$$\text{ = }\frac{6\text{ }\pm\text{ }\sqrt[]{1616}}{-10}$$\text{ t}_1\text{ = }\frac{6\text{ + }40.1995}{-10}\text{ = }\frac{46.1995}{-10}\text{ = -4.61995 }\approx\text{ -4.62 seconds}$$undefined$