91Ó°ÊÓ

Free Throw Under certain conditions the maximum height \(y\) attained by a basketball released from a height \(h\) at an angle \(\theta\) measured from the horizontal with an initial velocity \(v_{0}\) is given by \(y=h+\left(v_{0}^{2} \sin ^{2} \theta\right) / 2 g\), where \(g\) is the acceleration due to gravity. Compute the maximum height reached by a free throw if \(h=\) \(2.15 \mathrm{~m}, v_{\mathrm{o}}=8 \mathrm{~m} / \mathrm{s}, \theta=64.47^{\circ}\), and \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\)

Use a calculator in radian mode to compare the values of \(\tan (1.57)\) and \(\tan (1.58)\). Explain the difference in these values.

Verify the given identity. $$ \cot 2 x=\frac{1}{2}(\cot x-\tan x) $$

Verify the given identity. $$ (1-\tan \beta)^{2}(1+\tan \beta)^{2}+4 \tan ^{2} \beta=\sec ^{4} \beta $$

Verify that \(\operatorname{arccot} x=\frac{\pi}{2}-\arctan x,\) for all real numbers \(x\).

Find the first three \(x\) -intercepts of the graph of the given function on the positive \(x\) -axis. $$ f(x)=1+\cos \pi x $$

Use a calculator in radian mode to compare the values of \(\cot (3.14)\) and \(\cot (3.15)\)

Putting the Shot The range of a shot put released from a height \(h\) above the ground with an initial velocity \(v_{0}\) at an angle \(\theta\) to the horizontal can be approximated by: $$ R=\frac{v_{0} \cos \theta}{g}\left(v_{0} \sin \theta+\sqrt{v_{0}^{2} \sin ^{2} \theta+2 g h}\right) $$ where \(g\) is the acceleration due to gravity. If \(v_{\mathrm{o}}=\) \(13.7 \mathrm{~m} / \mathrm{s}, \theta=40^{\circ},\) and \(g=9.81 \mathrm{~m} / \mathrm{s}^{2},\) compare the ranges achieved for the release heights (a) \(h=2.0\) \(\mathrm{m}\) and (b) \(h=2.4 \mathrm{~m} .\) (c) Explain why an increase in \(h\) yields an increase in \(R\) if the other parameters are held fixed. (d) What does this imply about the advantage that height gives a shot-putter?

Graphically verify the given identity. $$ \sin (x+\pi)=-\sin x $$

Verify the given identity. $$ \sec 2 x=\frac{1}{2 \cos ^{2} x-1} $$