please solve the problem.If a person drops a pebble in a well of 100 m depth,after how much time will he be able to hear the sound starting from the time he drops the pebble (taking g=10ms-2 and speed of sound=343ms-1).
Answers
To calculate the total time after which the sound of the pebble hitting the surface of the well reaches the person, we need to calculate the time taken by the pebble to reach the bottom and time taken by the sound waves to reach the top.
Given initial velocity,u=
,depth of the well,s=100 m and acceleration, 
Let t¹ be the time taken by the pebble to hit the bottom.
From second equation of motion,
s=ut+
100=(0×t)+(×10×t²)
=>100=5
Given, speed of sound=
and distance covered=100 m
Therefore, time for sound to reach the top,
=>
=
=0.3 seconds (approximate)
Therefore, total time,t =
+=
Answer:
Solve the following inequality :
\displaystyle{\dfrac{1-\sqrt{21-4x-x^2}}{x+1}\geq 0}
x+1
1−
21−4x−x
2
≥0{({({2}^{\sqrt{x}\times 3})}^{\frac{1}{2}\sqrt{x}})}^{\frac{2}{\sqrt{x-1}}}=2((2
√
x
×3
)
2
1
√
x
)
√
x−1
2
=2
1 Regroup terms.
{({({2}^{3\sqrt{x}})}^{\frac{1}{2}\sqrt{x}})}^{\frac{2}{\sqrt{x-1}}}=2((2
3√
x
)
2
1
√
x
)
√
x−1
2
=2
2 Simplify \frac{1}{2}\sqrt{x}
2
1
√
x
to \frac{\sqrt{x}}{2}
2
√
x
.
{({({2}^{3\sqrt{x}})}^{\frac{\sqrt{x}}{2}})}^{\frac{2}{\sqrt{x-1}}}=2((2
3√
x
)
2
√
x
)
√
x−1
2
=2
3 Use Power Rule: {({x}^{a})}^{b}={x}^{ab}(x
a
)
b
=x
ab
.
{({2}^{3\sqrt{x}\times \frac{\sqrt{x}}{2}})}^{\frac{2}{\sqrt{x-1}}}=2(2
3√
x
×
2
√
x
)
√
x−1
2
=2
4 Use Quotient Rule: \frac{{x}^{a}}{{x}^{b}}={x}^{a-b}
x
b
x
a
=x
a−b
.
\[{({2}^{\frac{3\sqrt[2}+\frac{1}{2]{x}}{2}})}^{\frac{2}{\sqrt{x-1}}}=2\]
5 Simplify \frac{1}{2}+\frac{1}{2}
2
1
+
2
1
to 11.
{({2}^{\frac{3{x}^{1}}{2}})}^{\frac{2}{\sqrt{x-1}}}=2(2
2
3x
1
)
√
x−1
2
=2
6 Use Rule of One: {x}^{1}=xx
1
=x.
{({2}^{\frac{3x}{2}})}^{\frac{2}{\sqrt{x-1}}}=2(2
2
3x
)
√
x−1
2
=2
7 Use Power Rule: {({x}^{a})}^{b}={x}^{ab}(x
a
)
b
=x
ab
.
{2}^{\frac{6x}{2\sqrt{x-1}}}=22
2√
x−1
6x
=2
8 Simplify \frac{6x}{2\sqrt{x-1}}
2√
x−1
6x
to \frac{3x}{\sqrt{x-1}}
√
x−1
3x
.
{2}^{\frac{3x}{\sqrt{x-1}}}=22
√
x−1
3x
=2
9 Use Definition of Common Logarithm: {b}^{a}=xb
a
=x if and only if log_b(x)=alog
b
(x)=a.
\frac{3x}{\sqrt{x-1}}=\log_{2}{2}
√
x−1
3x
=log
2
2
10 Use Change of Base Rule: \log_{b}{x}=\frac{\log_{a}{x}}{\log_{a}{b}}log
b
x=
log
a
b
log
a
x
.
\frac{3x}{\sqrt{x-1}}=\frac{\log{2}}{\log{2}}
√
x−1
3x
=
log2
log2
11 Simplify \frac{\log{2}}{\log{2}}
log2
log2
to 11.
\frac{3x}{\sqrt{x-1}}=1
√
x−1
3x
=1
12 Square both sides
\frac{9{x}^{2}}{x-1}=1
x−1
9x
2
=1
13 Multiply both sides by x-1x−1.
9{x}^{2}=x-19x
2
=x−1
14 Move all terms to one side.
9{x}^{2}-x+1=09x
2
−x+1=0
15 Use the Quadratic Formula.
1 In general, given a{x}^{2}+bx+c=0ax
2
+bx+c=0, there exists two solutions where:
x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}x=
2a
−b+√
b
2
−4ac
,
2a
−b−√
b
2
−4ac
2 In this case, a=9a=9, b=-1b=−1 and c=1c=1.
{x}^{}=\frac{1+\sqrt{{(-1)}^{2}-4\times 9}}{2\times 9},\frac{1-\sqrt{{(-1)}^{2}-4\times 9}}{2\times 9}x
=
2×9
1+√
(−1)
2
−4×9
,
2×9
1−√
(−1)
2
−4×9
3 Simplify.
x=\frac{1+\sqrt{35}\imath }{18},\frac{1-\sqrt{35}\imath }{18}x=
18
1+√
35
ı
,
18
1−√
35
ı
x=\frac{1+\sqrt{35}\imath }{18},\frac{1-\sqrt{35}\imath }{18}x=
18
1+√
35
ı
,
18
1−√
35
ı
Done
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\frac{1-\sqrt{21-4x}-{x}^{2}}{x+1}\ge 0
x+1
1−√
21−4x
−x
2
≥0
1 Replace the inequality sign with an equal sign,
so that we can solve it like a normal equation.
\frac{1-\sqrt{21-4x}-{x}^{2}}{x+1}=0
x+1
1−√
21−4x
−x
2
=0
2 Multiply both sides by x+1x+1.
1-\sqrt{21-4x}-{x}^{2}=01−√
21−4x
−x
2
=0
3 Separate terms with roots from terms without roots.
-\sqrt{21-4x}=-1+{x}^{2}−√
21−4x
=−1+x
2
4 Regroup terms.
-\sqrt{21-4x}={x}^{2}-1−√
21−4x
=x
2
−1
5 Square both sides
21-4x={(x+1)}^{2}{(x-1)}^{2}21−4x=(x+1)
2
(x−1)
2
6 Expand.
21-4x={x}^{4}-2{x}^{3}+{x}^{2}+2{x}^{3}-4{x}^{2}+2x+{x}^{2}-2x+121−4x=x
4
−2x
3
+x
2
+2x
3
−4x
2
+2x+x
2
−2x+1
7 Simplify {x}^{4}-2{x}^{3}+{x}^{2}+2{x}^{3}-4{x}^{2}+2x+{x}^{2}-2x+1x
4
−2x
3
+x
2
+2x
3
−4x
2
+2x+x
2
−2x+1 to {x}^{4}-2{x}^{2}+1x
4
−2x
2
+1.
21-4x={x}^{4}-2{x}^{2}+121−4x=x
4
−2x
2
+1
8 Move all terms to one side.
21-4x-{x}^{4}+2{x}^{2}-1=021−4x−x
4
+2x
2
−1=0
9 Simplify 21-4x-{x}^{4}+2{x}^{2}-121−4x−x
4
+2x
2
−1 to 20-4x-{x}^{4}+2{x}^{2}20−4x−x
4
+2x
2
.
20-4x-{x}^{4}+2{x}^{2}=020−4x−x
4
+2x
2
=0
10 No root was found algebraically. However, the following root(s) were found by numerical methods.
x=-2.567480,2.129130x=−2.567480,2.129130
11 Also note that xx is undefined at -1−1.
x\ne -1x≠−1
12 From the values of xx above, we have these 4 intervals to test.
\begin{aligned}&x\le -2.567480\\&-2.567480\le x\le -1\\&-1\le x\le 2.129130\\&x\ge 2.129130\end{aligned}
x≤−2.567480
−2.567480≤x≤−1
−1≤x≤2.129130
x≥2.129130
13 Pick a test point for each interval.
For the interval x\le -2.567480x≤−2.567480:
Let's pick x=-3x=−3. Then, \frac{1-\sqrt{21-4\times -3}-{(-3)}^{2}}{-3+1}\ge 0
−3+1
1−√
21−4×−3
−(−3)
2
≥0.
After simplifying, we get 6.872281\ge 06.872281≥0, which is true.
Keep this interval..
For the interval -2.567480\le x\le -1−2.567480≤x≤−1:
Let's pick x=-2x=−2. Then, \frac{1-\sqrt{21-4\times -2}-{(-2)}^{2}}{-2+1}\ge 0
−2+1
1−√
21−4×−2
−(−2)
2
≥0.
After simplifying, we get 8.385165\ge 08.385165≥0, which is true.
Keep this interval..
For the interval -1\le x\le 2.129130−1≤x≤2.129130:
Let's pick x=0x=0. Then, \frac{1-\sqrt{21-4\times 0}-{0}^{2}}{0+1}\ge 0
0+1
1−√
21−4×0
−0
2
≥0.
After simplifying, we get -3.582576\ge 0−3.582576≥0, which is false.
Drop this interval..
For the interval x\ge 2.129130x≥2.129130:
Let's pick x=3x=3. Then, \frac{1-\sqrt{21-4\times 3}-{3}^{2}}{3+1}\ge 0
3+1
1−√
21−4×3
−3
2
≥0.
simplifying, we get -2.75\ge 0−2.75≥0, which is false.