The acceleration due to gravity on the moon is -5.3 ft/sec/sec. When your Space

Question

The acceleration due to gravity on the moon is -5.3 ft/sec/sec. When your Space X trip arrives on the moon, you decide to throw your NCC hat directly upward in celebration. If you release the top of the stairway which is 13.25 ft above the surface of the moon with an upward velocity of 21.2 ft/sec, please determine: The equation of the velocity of the hat (as a function of time). The equation of the height of the hat (as a function of time). The maximum height of the hat. When the hat comes back to the starting point. When the hat reaches the surface of the moon.

Explanation / Answer

acceleration a(t)=-5.3

initial height h(0)=13.25

initial velocity v(0)=21.2

a)

velocity v(t)=a(t) dt

v(t)=(-5.3) dt

v(t)=-5.3t +c

v(0)=21.2

(-5.3*0) +c=21.2

0+c=21.2

c=21.2

v(t)=-5.3t +21.2

b)

height h(t)=v(t) dt

h(t)=(-5.3t +21.2) dt

h(t)=(-5.3/2)t2 +21.2t +c

h(t)=-2.65t2 +21.2t +c

h(0)=13.25

(-2.65*02)+(21.2*0) +c=13.25

c=13.25

h(t)=-2.65t2 +21.2t +13.25

c) for maximum height h'(t)=v(t)=0

-5.3t +21.2=0

t=21.2/5.3

t=4

maximum height =h(4)

maximum height =(-2.65*42)+(21.2*4) +13.25

maximum height =55.65 ft

d)at starting point h(t)=13.25

-2.65t2 +21.2t +13.25=13.25

2.65t2 =21.2t

t =21.2/2.65

t=8

hat comes back to starting point after 8 sec

e)

when hat reaches surface of moon ,h(t)=0

-2.65t2 +21.2t +13.25=0

by quadratic formula

t=-0.582576,8.58258

t cannot be negative

so

after 8.58 sec hat reaches surface of moon

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