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An astronaut shipwrecked on a distant planet with unknown characteristics is on

ID: 1334023 • Letter: A

Question

An astronaut shipwrecked on a distant planet with unknown characteristics is on top of a cliff, which he wishes to descend. He does not know the acceleration due to gravity on the planet, and he has only a good watch with which to make measurements. He wants to learn the height of the cliff, and to do this he makes two measurements. First, he lets a rock fall from rest off the cliff edge; he finds that the rock takes 4.00 s to reach the distant ground. Second, he releases the rock from the same spot but tosses it upward so that it rises a height of what he estimates to be 2 m before it falls to the ground below. This time the rock takes 5.00 s to reach the ground. What is the height of the cliff?

Explanation / Answer

let height is h and acceleration due to gravity g.

in first experiment,

0.5*g*4^2=h

==>g=h/8...(1)

in second experiemnt:

let speed of tossing is v.

maximum height reached=v^2/(2*g)

it is given that height is 2 m

==>v^2=4*g

==>v=2*sqrt(g)

now, taking donward direction direction as positive and using the formula:

displacement=intiail veloicty*time+0.5*acceleration*time^2

==>h=-v*5+0.5*g*25

==>8*g=-10*sqrt(g)+12.5*g

==>10*sqrt(g)=4.5*g

==>g=4.9383 m/s^2

then h=8*g=39.5 m

hence the height of the cliff is 39.5 m

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