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DerKrebs [107]
4 years ago
5

A rock climber throws a small first aid kit to another climber who is higher up the mountain. The initial velocity of the kit is

5.70 m/s at an angle of 45.7 ° above the horizontal. At the instant when the kit is caught, it is traveling horizontally, so its vertical speed is zero. What is the vertical height between the two climbers?
Physics
1 answer:
mariarad [96]4 years ago
5 0

Answer:

Height difference between climbers = 0.85 m.

Explanation:

<u>Vertical motion of first aid kit:</u>

Initial vertical velocity, u = 5.70 sin 45.7 = 4.08 m/s

Acceleration, g = -9.81 m/s²

When the climber who is higher up the mountain catches the kit it's vertical speed is zero.

Final vertical velocity, v = 0 m/s

We have equation of motion, v² = u² + 2as

Substituting

            0² = 4.08² + 2 x -9.81 x s      

            s = 0.85 m

So the kit travels 0.85 m vertically.

Height difference between climbers = 0.85 m.

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<h2>Answer:</h2>

He saves 13.2 minutes

<h2>Explanation:</h2>

Hey! The question is incomplete, but it can be found on the internet. The question is:

How many minutes did he save?

Let's call:

t_{1}:Time \ at \ speed \ 65mph \\ \\ t_{2}:Time \ at \ speed \ 73mph \\ \\ v_{1}=65mph \\ \\ v_{2}=73mph

We know that the 135 miles are on the interstate highway where the speed limit is 65 mph. From this, we can calculate the time it takes to drive on this highway. Assuming Richard maintains constant the speed:

v=\frac{d}{t} \\ \\ d:distance \\ \\ t:time \\ \\ v:velocity \\ \\ t_{1}=\frac{d}{v_{1}} \\ \\ t=\frac{135}{65} \\ \\ t_{1}=2.07 \ hours

Today he is running late and decides to take his chances by driving at 73 mph, so the new time it takes to take the trip is:

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