# Calculus Hw?

a stone was thrown upward from the surface of Titan (the largest moon of the planet saturn) with velocity 8 meters/sec. find the maximum height (round to two decimal places) of the stone given that the acceleration due to gravity on the surface of Titan is 235.1 meters/sec^2.

Show steps please I am very confused!

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• Let the height of the stone at time t be y m from the surface

of Titan, then

y"=-235.1

=>

y'd(y')/dy=-235.1

=>

[(y')^2]/2=-235.1y+C

y=0, y'=8

=>

32=C

=>

[(y')^2]/2=32-235.1y

When the stone reaches the

max. height, y'=0

=>

max.y=32/235.1=0.1361 m/s

approximately.

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• Is this really the problem statement? 235.1 m/s^2 is nearly 24 times Earth's surface gravity. No planet has a gravitational acceleration anywhere near that. The strongest acceleration (probably well below the surface) is a little over one-tenth of that.

Anyway, to solve height in terms of velocity, equate the kinetic energy to the work that energy need to do to raise a mass m by a height h with a gravitational acceleration g:

(1/2)mv^2 = mgh . . . . energy = work

v^2 = 2gh . . . . divide by m to cancel, multiply by 2 to get rid of the fraction

h = v^2 / (2g)

That's it. Plug numbers to get h = 8^2 / (2 * 235.1) = 64 / 470.2 = 0.136 m. Much simpler than solving for elapsed time first.

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• a(t) = -235.1

v(t) = -235.1 * t + C

v(0) = 8

8 = -235.1 * 0 + C

8 = C

v(t) = -235.1 * t + 8

s(t) = -117.55 * t^2 + 8t + C

s(0) = 0

s(t) = -117.55 * t^2 + 8t + 0

s(t) = 8t - 117.55 * t^2

v(t) = 0. This tells you when we're at a maximum height.

v(t) = -235.1 * t + 8

0 = 8 - 235.1 * t

0 = 80 - 2351 * t

2351 * t = 80

t = 80 / 2351

Plug that value for t in to s(t)

s(t) = 8t - 117.55 * t^2

s(80/2351) = 8 * (80/2351) - 117.55 * (80/2351)^2

s(80/2351) = (80/2351)^2 * (8 * 2351/80 - 117.55)

s(80/2351) = (80/2351)^2 * (235.1 - 117.55)

s(80/2351) = (80/2351)^2 * (117.55)

s(80/2351) = 0.13611229264142917907273500638026....

It'll achieve a maximum height of 13.61 cm

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