When will the rock hit the surface and with what velocity?

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SUMMARY

The discussion focuses on the motion of a rock thrown upward on Mars with an initial velocity of 10 m/s, described by the height equation H = 10t - 1.86t². The velocity of the rock at time t is given by the derivative v = 10 - 3.72t. The rock hits the surface at approximately t = 5.37 seconds, with a velocity of -9.9764 m/s upon impact, indicating it descends at nearly the same speed it was thrown upward.

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Homework Statement


"If a rock is thrown upward on the planet Mars with a velocity of 10 m/s, its height (in meters) after t seconds is given by H = 10t - 1.86t^2."

a) Find the velocity of the rock when t = a
b) Find the velocity of the rock after 1 second
c) When will the rock hit the surface?
d) With what velocity will the rock hit the surface?

The Attempt at a Solution



I was able to do parts a and b. I am stuck at c and d though.

For (a) I derived the formula that is given.
Which turns to: 10 - 3.72a
For (b) I substituted 1 into the equation which gives: 6.28

Now how would I go about finding out when the rock will hit the surface and it's velocity? Any help appreciated.
 
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your velocity is the derivative... v= 10-3.72t

there are going to be 2 times when the rock is on the ground.
one at time zero and one at the 2nd time... which is 10/1.86t

1.86t2-10t=0
1.86t=10
t=0 and t= (about) 5.37

to find the velocity

you plug in 5.37 into the derivative:

v= 10-3.72t
= 10-3.72(5.37)= -9.9764

which equals... negative 10
as we can see, it was thrown up at 10 m/s and coming down it should be the opposite of that... -10.

:) enjoy!

need any more explanations?
 

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