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 √(c^2-a^2)=b \qquad \begin{cases} c=\stackrel{10}{hypotenuse}\\ a=\stackrel{9.25}{adjacent}\\ b=opposite\\ \end{cases} \\\\\\ √(10^2-9.25^2)=b\implies √(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

a) average speed = distance / time

=
`(10)/(1.6)`

= 6.25 ft/s = 6.3 ft/s

b) average horizontal speed

= v cos θ

=
`6.3 * (9.25)/(10)`

= 5.875 ft/s = 5.9 ft/s

c) average vertical speed

= v sin θ

=
`6.3 * \frac{\sqrtr{10^2-9.25^2}}{10}`

= 2.393 ft/s = 2.4 ft/s