Question

the altitude of an object, d, can be modeled using the equation d=-16t^2+vt+h. a cannon sits on top of a hill that is 396 feet high. A cannonball is fired with an initial upward velocity of 100 ft./s how long before the cannon ball strikes the ground

Answer 1

Answer:

9 seconds

Step-by-step explanation:

We are given $v=100$ ($v$ is the initial velocity) and $h=396$ ($h$ is the initial height) in the equation:

$d=-16t^2+vt+h$.

So we have:

$d=-16t^2+100t+396$.

We want to know when the attitude has reached 0. We will go with the later time since we are looking for when it has hit the ground.

We are solving the equation:

$0=-16t^2+100t+396$

$-16t^2+100t+396=0$

Divide both sides by 4:

$-4t^2+25t+99=0$

I'm going to use the quadratic formula.

Let's compare your equation to $ax^2+bx+c=0$ to find the values we will use for $a,b,\text{and} c$.

$a=-4$

$b=25$

$c=99$

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

$x=\frac{-25 \pm \sqrt{25^2-4(-4)(99)}}{2(-4)}$

$x=\frac{-25 \pm \sqrt{2209}}{-8}$

We have either:

$x=\frac{-25+\sqrt{2209}}{-8} \text{ or } \frac{-25-\sqrt{2209}}{-8}$

$x=-2.75 or 9$

So when t=9 seconds is when the cannon has hit the ground.