jungle Jim runs along level ground at 5 m/s. He grabs a vine and swings up to the edge of a nearby cliff (above the level ground on which he was running), just making it to the height of thecliff, essentially at rest. He then steps off onto the cliff.1. How high is the cliff above the level ground? Jungle Jim then runs at 5 m/s and jumps off the cliff, back to the level ground below. What is his speed just before hitting the level ground (ignore air resistance)?

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aristocles aristocles · Answer 1 · 2020-04-06T20:22:36+02:00

Answer:

Part a)

Height of the cliff is given is 1.27 m

Part b)

Speed by which he land on the ground is 7.07 m/s

Explanation:

Part a)

As we know that Jim is initially running with speed 5 m/s

so here after holding the vine it will curl into the circle and reach the top where his whole kinetic energy will convert into potential energy

So here we have

[tex]KE = U[/tex]

[tex]\frac{1}{2}mv^2 = mgH[/tex]

[tex]H = \frac{v^2}{2g}[/tex]

[tex]H = \frac{5^2}{2(9.81)}[/tex]

[tex]H = 1.27 m[/tex]

Part b)

Now again he jumps from the cliff and land on the ground

so again by work energy theorem

[tex]K_i + mgH = K_f[/tex]

so we have

[tex]\frac{1}{2}mv_i^2 + mgH = \frac{1}{2}mv_f^2[/tex]

[tex]v_f = \sqrt{v_i^2 + 2gH}[/tex]

[tex]v_f = \sqrt{5^2 + 2(9.8)(1.27)}[/tex]

[tex]v_f = 7.07 m/s[/tex]