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

It takes approximately 3 seconds for the ball to hit the ground after being thrown.

Step-by-step explanation:

To determine the time it takes for the ball to hit the ground, we need to find when the height function H becomes zero. In the given equation H = 79 - 6t - 5t^2, set H to zero and solve for t. This quadratic equation represents the ball's height as a function of time.

By factoring or using the quadratic formula, we find two solutions for t. However, we discard the negative solution since time cannot be negative in this context. The positive solution gives us the time it takes for the ball to hit the ground.

The equation models the ball's height as it falls due to gravity, with the initial downward velocity of 6 m/s and the gravitational acceleration represented by the term -5t^2. The initial height of 79 meters is also considered in the equation.

In summary, by setting the height function to zero and solving for the positive root of the resulting quadratic equation, we find that the ball hits the ground approximately 3 seconds after being thrown.

Answer 2

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{ = }(46.1995)/(-10)\text{ = -4.61995 }\approx\text{ -4.62 seconds}`