A falling body dropped at a velocity
\(60\, \mathrm{m}\mathrm{s}^{-1}\). Find the
initial height \(h\),
if the relation between the velocity and the initial height
\(h\) is
\(v = \sqrt{2hg}\). Use
\(g = 10\, \mathrm{m}\mathrm{s}^{-2}\) for
acceleration of gravity.
The initial height is between \(150\, \mathrm{m}\)
and \(200\, \mathrm{m}\).
The initial height is smaller than \(100\, \mathrm{m}\).
The initial height is between \(100\, \mathrm{m}\)
and \(150\, \mathrm{m}\).
The initial height is bigger than \(200\, \mathrm{m}\).
Consider the equation
\[
\sqrt{x^{2 } - 2x + 1} = x + 2
\]
and the equation which arises from this equation by squaring both sides of the
equation, i.e. the equation
\[
\left (\sqrt{x^{2 } - 2x + 1}\right )^{2} = (x + 2)^{2}.
\]
Identify a true statement.
Both equations are equivalent only if
\(x\geq - 2\).
Both equations are equivalent.
Both equations are equivalent only if
\(x\leq - 2\).
Removing radical in an equation by squaring both sides may enrich the set of
solutions of this equation and checking the solutions of the new equation in the
original equation may be necessary. Identify a correct conclusion in the particular
case of the following equation.
\[
-\sqrt{x^{2 } - 2x + 1} = x
\]
If we look for the solution in the set
\(\mathbb{R}^{-}\), then
squaring both sides of the equation gives an equivalent equation. The checking of the
solution is not necessary.
If we look for the solution in the set
\(\mathbb{R}^{+}\), then
squaring both sides of the equation gives an equivalent equation. The checking of the
solution is not necessary.
If we look for the solution in the set
\(\mathbb{R}\), then
squaring both sides of the equation gives an equivalent equation. The checking of the
solution is not necessary.
A body hangs on a string of the length
\(l_{1}\). The length
\(l\) of the spring
defines the period \(T\)
of motion by the relation
\[
T = 2\pi \sqrt{ \frac{l}
{g}},
\]
where \(g\)
is a standard acceleration of gravity. We have to adjust the length of the string such
that the period doubles. Find the new length of the string.
We elongate the string by \(3\cdot l_{1}\),
i.e. \(l_{2} = l_{1} + 3l_{1}\).
The length doubles, i.e. \(l_{2} = 2l_{1}\).
The new length will be half of the original length, i.e.
\(l_{2} = \frac{1}
{2}l_1\).
We shorten the string by \(3\cdot l_{1}\),
i.e. \(l_{2} = l_{1} - 3l_{1}\).