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 1389 feet tall. Use the position function below for free-falling objects.

s(t) = −16t² + v₀t + s₀

(b) Determine the average velocity on the interval [1, 2].

________ ft/s

(c) Find the instantaneous velocities when t = 1 seconds and t = 2 seconds.

v(1) =  ________ft/s

v(2) =  ________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

In one second, it traveled -16 feet

In two seconds, it traveled -16(2)^2 = - 64 feet

from 16 to 64 is 48 feet in one second from t=1 to t=2

or average velocity of 48 feet per second

instantaneous velocity is the slope or derivative of the h(t) function or

f''(t) = -32t

f'(1) = -32 feet per second

f'(2) = -32(2) = -64 feet per second

to find when it reaches the ground set h(t) = 0

or

-16t^2 + 1389 = 0

16t^2 = 1389

t^2 = 1389/16 = 86.8125

t = sqr(86.8125) = 9.317

speed at impact =

h'(9.317) = -32(9.317) = -298.144 feet per second

Answer 2

initial position is 1389. initial velocity is 0 since it was dropped, not thrown.

s(t) = −16t² + v₀t + s_{0 becomes} s(t) = −16t² + 1389

find the position for t=1 and t=2. s(1) = 1373 and s(2) = 1325. these are the answers to part c.

in one second (from t=1 to t=2), the silver dollar traveled 1373-1325 = 48 ft. divide by the time difference (2-1 = 1) and that gives you the average velocity over the time interval...48 ft/sec

when the coin hits the ground, s(t) = 0.

solve −16t² + 1389 = 0. t² = 86.8125, so t = 9.3173225767921119024790410312156...

or approx 9.317

to get the instantaneous velocity at a given time t, take the derivative of the position/displacement function.

you get v(t) = -32t.

plug in the time it takes to hit the ground (from above) and you get v(t) = -32(9.317) = 298.144 ft/sec