A silver dollar is dropped from the top of a building that is 1380 feet tall. Us

Question

A silver dollar is dropped from the top of a building that is 1380 feet tall. Use the position function below for free-falling objects. s(t) = -16t^2 + V_0t + s_0 Determine the position and velocity functions for the coin. Determine the average velocity on the interval [3, 4] Find the instantaneous velocities when t = 3 seconds and t = 4 seconds. Find the time required for the coin to reach the ground level. (Round your answer to three decimal places.) Find the velocity of the coin at impact. (Round your answer to three decimal places.)

Explanation / Answer

Determine the position and velocity functions for the coin.

Since the coin was "dropped", the initial velocity is zero.
The initial height is 1380
So the position function is: s(t) = -16t^2 + 1380

Since v(t)= s'(t) , then the velocity function is : v(t)= -32t

B) Determine the average velocity on the interval [3,4].

Average velocity is the average change in s(t), so use slope formula: [ s(4)- s(3)]/ (4-3)

s(4) = -16(4)^2 + 1380 = 1124

s(3) = -16(3)^2 + 1380 = 1236

so           (1124-1236)/(4-3) = -112

C) Find the instantaneous velocities when t=3 and t=4.

Find V(3) and V(4)

v(3) = -32(3) = -96

v(4) = -32(4) = -128

D) Find the time required for the coin to reach ground level.

It reaches the ground when s(t)= 0

Solve : -16t^2 + 1380= 0, discarding negative values of t.

-16 t^2 = -1380

t^2 = 1380/16   =86.25

t = 9.287

E) Find the velocity of the coin at impact.

Find the value of v(t) for the value of t in part D

==== -32(9.287)

== -297.184

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