Question

30 points PLEASE HELP WITH MATH PROBLEM

Answer 1

10) What is the base of the triangle?

From this triangle we know:

THE AREA:

$A=80cm^2$

THE HEIGHT:

$H=B+12$

We know that the area, height, and base of a triangle are related according to the following formula:

$A=\frac{BH}{2} \\ \\ Substituting \ H: \\ \\ A=\frac{B(B+12)}{2} \\ \\ Substituting \ A: \\ \\  80=\frac{B(B+12)}{2} \\ \\ 2(80)=B(B+12) \\ \\ 160=B^2+12B \\ \\ B^2+12B-160=0$

Solving B by quadratic formula:

$B_{12}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ B_{12}=\frac{-12 \pm \sqrt{12^2-4(1)(-160)}}{2(1)} \\ \\ B_{12}=\frac{-12 \pm \sqrt{12^2-4(1)(-160)}}{2(1)} \\ \\ B_{1}=8 \ and \ B_{2}=-20$

Since we can’t have a negative value of the base, the correct option is:

$B=8cm$

11) How long will it take the water balloon to hit the ground?

We must use the formula:

$h(t)=-5.2t^2+v_{0}t+h_{0}$

Since James drops the balloon from a height of 45m, then this is the initial height, so $h_{0}=45m$. Moreover, at the very instant he drops the balloon the initial velocity is zero, so $v_{0}=0$. When the ballon hit the ground $h(t)=0$. Therefore:

$0=-5.2t^2+45$

Solving this equation:

$5.2t^2=45 \\ \\ t^2=\frac{45}{5.2} \\ \\ t^2=8.653 \\ \\ t=\sqrt{8.653} \\ \\ t=2.941s$

Rounding to the nearest tenth:

$\boxed{t=2.9s}$

Finally, the water balloon will hit the ground after 2.9 seconds.

12) Height and initial velocity

We have the equation:

$h(t)=-16t^2+32t+12$

But we know that this equation follows the form:

$h(t)=-16t^2+v_{0}t+h_{0}$

According to this:

The height of the platform is $h_{0}=12ft$

The initial velocity of the ball is $v_{0}=32ft/s$