Posted by **Anonymous** on Thursday, August 11, 2011 at 6:36am.

a key is dropped from a height of 30 m , If its distance from the ground at t seconds after the release is given by the position function d (t) = 30-5tÂ², find

a.its height velocity and acceleration after 1 second.

b. the time it reached the ground.

c. its position after 1 second

- calculus -
**Reiny**, Thursday, August 11, 2011 at 7:51am
If d(t) = 30 - 5t^2 m, then the velocity is

v(t) = -10t m/sec , and the acceleration is

a(t) = -10 m/sec^2

(each is the derivative of the previous function)

a) when t = 1

distance = 30 - 5(1^2) = 25 m

velocity = -10(1) = -10 m/s

acceleration = -10 m/s^2

b) when it reaches the ground, d(t) = 0, so

0 = 30 - 5t^2

t^2 = 6

t = √6 or appr. 2.4 seconds

c) 25 m above the ground , already answered in a)

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