Question

Find the average rate of horizontal change of the ball the 1.6 sec.

Answer 1

a)

the average speed, well it does 10 ft in 1.6 secs, so is simply their quotient, 10/1.6 = 6.25 ft/s on average.

b)

over the horizontal, well, it covered horizontally 9.25 ft in those 1.6 secs, so their quotient is 9.25/1.6 = 5.78125 ft/s on average.

c)

we don't know the vertical distance, but

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=\stackrel{10}{hypotenuse}\\ a=\stackrel{9.25}{adjacent}\\ b=opposite\\ \end{cases} \\\\\\ \sqrt{10^2-9.25^2}=b\implies \sqrt{14.4375}=b\implies 3.8\approx b$

so, it covered vertically about 3.8 ft in those 1.6 secs, so then 3.8/1.6 = 2.375 ft/s on average

Answer 2

Answer:

a) average speed = distance / time

                              = $\dfrac{10}{1.6}$

                              = 6.25 ft/s  = 6.3 ft/s

b) average horizontal speed

                                   = v cos  θ

                                   = $6.3 \times \dfrac{9.25}{10}$

                                   = 5.875 ft/s  = 5.9 ft/s

c) average vertical  speed

                                   = v sin  θ

                                   = $6.3 \times \dfrac{\sqrtr{10^2-9.25^2}}{10}$

                                   = 2.393 ft/s  = 2.4 ft/s