A girl ties a rope securely to the windowsill of a tower and climbs down to the base of it. She is with a group of friends and wants them to come down too, but easier. She ties the other end of the rope to a tree 30m down and 60m away horizontally, creating a zipline. The first person down the zip line weighs 80 kg. How quickly does he accelerate, and how long does it take for him to reach the end of the zipline?

[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \text{tan}\theta=\frac{60}{20} \\ \tan \theta=2 \\ \theta=tan^{-1}(2)=63.43 \end{gathered}[/tex]

Then, the net force in the direction of the rope is equal to:

[tex]\begin{gathered} F_{\text{net}}=mg\cos \theta \\ F_{\text{net}}=(80\operatorname{kg})(9.8m/s^2)\cos (63.43) \\ F_{\text{net}}=350.62N \end{gathered}[/tex]

By the second law of Newton, this force is equal to mass times acceleration, so we can solve for acceleration as follows:

[tex]\begin{gathered} F_{\text{net}}=ma \\ a=\frac{F_{net}}{m}=\frac{350.62N}{80\operatorname{kg}}=4.38m/s^2 \end{gathered}[/tex]

So, the first person accelerates at 4.38 m/s².

Now, we need to find the length of the rope. Using the Pythagorean theorem, we get:

[tex]\begin{gathered} L=\sqrt[]{60^2+30^2} \\ L=\sqrt[]{3600+900} \\ L=\sqrt[]{4500}=67.08\text{ m} \end{gathered}[/tex]

Then, using a kinetic equation, we get:

[tex]\begin{gathered} x=v_it+\frac{1}{2}at^2 \\ x=\frac{1}{2}at^2 \\ 2x=at^2 \\ \frac{2x}{a}=t^2 \\ t=\sqrt[]{\frac{2x}{a}} \end{gathered}[/tex]

Where x is the distance traveled, vi is the initial velocity, which is 0 m/s, a is the acceleration and t is the time.

Now, we can replace x by the length of the rope 67.08m and a by 4.38 to get:

[tex]t=\sqrt[]{\frac{2(67.08)}{4.38}}=5.54\text{ s}[/tex]

So, the first person takes 5.54 s to reach the end of the zipline.

Therefore, the answers are

Acceleration = 4.38 m/s²

Time = 5.54 s