Posted by **Linda** on Saturday, June 8, 2013 at 10:12pm.

If an arrow is shot straight up from the surface of the moon with an initial velocity of 100 ft/s, its height in feet after t second is given by s(t)=(100t)−(83/100)(t^2).

Use the limit definition of the derivative to find the answers to the following questions.

Find the velocity of the arrow when t=2.

Find the velocity of the arrow when t=a.

When will the arrow return to the moon's surface?

What will the velocity of the arrow be when it hits the surface?

- Calculus -
**Reiny**, Saturday, June 8, 2013 at 10:53pm
s(2) = 100(2) - (83/100)(4) = 4917/25

s(2+h) = 100(2+h) - (83/100)(2+h)^2

= 200+100h - 332/100 - (332/100)h - (83/100)h^2

velocity = Limit ( s(2+h) - s(2) )/(2+h - 2) as h -->0

= Lim (200+100h - 332/100 - (332/100)h - (83/100)h^2 - 4917/25)/h

= Lim ( (2417/25)h - (83/100)h^2 )/h

= lim 2417/25 - (83/100)h , as h ---> 0

= 2417/25 ft/sec

carefully repeat the above steps using a instead of 2

you should get

v(a) = 100 - (83/50) a

the arrow will return to the surface when s(t) = 0

100t - (83/100)t^2 = 0

t (100 - (83/100)t ) = 0

t= 0 ----> the start of the shot

or

t = 100(100)/83 = appr 120.5 seconds

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