a silver dollar is dropped from the top of a building

Question

A silver dollar is dropped from the top of a building that is 1377 feet tall. Use the position function below for free-falling objects.

s(t) = −16t² + v₀t + s₀

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

________ ft/s

(c) Find the instantaneous velocities when t = 3 seconds and t = 4 seconds.

v(3) =  ________ft/s

v(4) =  ________ft/s

(d) Find the time required for the coin to reach the ground level. (Round your answer to three decimal places.)

t =  ________s

(e) Find the velocity of the coin at impact. (Round your answer to three decimal places.)

________ ft/s

Answer 1

Substitute v₀ = 0 and s₀ = 1377 into the position function provided:

s(t) = −16t² + v₀t + s₀

which becomes:

(1) s(t) = −16t² + 1377

The velocity function is the first derivative of (1), which is:

(2) v(t) = ds/dt = s'(t) = -32t

This is a linear function, so:

(b) The average velocity on the time interval [3,4] is the average of the velocity at the end points:

( v(3)+v(4) )/2 = ( -96 + -128 )/2 = - 112 ft/s

The answers to (c) are used in (b):

v(3) = -96 ft/s

v(4) = -128 ft/s

(d) When the coin reaches the ground level, s(t) = 0. Solve (1) for s(t) = 0.

0 = −16t² + 1377

16t² = 1377

t² = 1377/16

t = 9.277 s

(e) From (d), we know that it takes 9.277 seconds for the coin to reach the ground level. Substitute this value into (2) to find the velocity of the coin at impact:

v(9.277) = −32 (9.277) = -296.864 ft/s



Evaluate (1) for t = 3 and t = 4

s(3) = 1377 - 144 = 1233

s(4) = 1377 - 256 = 1121