A ball is kicked upward with an initial velocity of 52 feet per second. The ball's height, h (in feet), from the ground Is

Mathematics

1 answer:

Norma-Jean [14]2 years ago

5 0

Check the picture below.

$\textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+52}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 52}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (52)^2}{4(-16)}\right) \implies \left( - \cfrac{ 52 }{ -32 }~~,~~0 - \cfrac{ 2704 }{ -64 } \right)$

$\left( \cfrac{13}{8}~~,~~\cfrac{169}{4} \right)\implies \underset{\stackrel{\uparrow \hspace{3em}}{seconds\qquad }}{\stackrel{\stackrel{\qquad feet}{\hspace{3em}\downarrow }}{\left( 1\frac{5}{8}~~,~~42\frac{1}{4} \right)}}$

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If 1/√a-√b=1/3 and 1/√a+√b=1/2, then find the difference of a and b.

kow [346]

ANSWER:

If $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ and $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ then the difference of a and b is 6

SOLUTION:

Given, $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ → $\sqrt{a}-\sqrt{b}=3$ ----- (1)

And $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ → $\sqrt{a}+\sqrt{b}=2$ --- (2)

We have to find difference of a and b.

Now, add (1) and (2)

$\sqrt{a}-\sqrt{b}=3$

$\sqrt{a}+\sqrt{b}=2$

Adding above two equations, we get,

$2 \sqrt{a}+0=2+3$

$\begin{array}{l}{2 \sqrt{a}=5} \\\\ {\sqrt{a}=\frac{5}{2}} \\\\ {a=\frac{25}{4}}\end{array}$

substitute $\sqrt{a}$ value in (2)

$\begin{array}{l}{\frac{5}{2}+\sqrt{b}=2} \\\\ {\sqrt{b}=\frac{2}{\sin \frac{5}{2}}} \\\\ {\sqrt{b}=\frac{4-5}{2}} \\\\ {\sqrt{b}=\frac{-1}{2}} \\\\ {b=\frac{1}{4}}\end{array}$