A silver dollar is dropped from the top of a building that is 1351 feet tall. Use the position function below for free-falling objects.

Question

s(t) = −16t² + v₀t + s₀

(a) Determine the position and velocity functions for the coin.

s(t) =	−16t2+1351
v(t) =

(b) Determine the average velocity on the interval [2, 3].

 ft/s

(c) Find the instantaneous velocities when t= 2 seconds and  t= 3 seconds.

v(2) =  ft/s

v(3) =  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

Hi Shannon,

a)

The object is free-falling from 1,351 ft, which means that the initial velocity is 0 and the initial position is 1,351.

Then we get s(t) = -16t² + 0t +1351, which you got here.

Now, the velocity is the derivative of the position equation. (since it is the directional rate of change)

s'(t) = v(t) = -32t (Power Rule)

+Extra : Acceleration = v'(t) = -32 << which makes sense since the known gravity is 9.81m/s² ≈ 32 ft/s²

b)

To find the average velocity, connect two points on the position equation and find the slope of that line.

Here you need to find the avg. velocity on the interval [2,3].

This gives you the two endpoints, (2, 1287), (3, 1207).

Plugging these two points into the slope formula. (1287 - 1207) / (2 - 3) = - 80.

(make sure you include the negative sign. velocity is a vector)

c)

v(2) = - 64 ft/s

v(3) = - 96 ft/s

d)

You need to find t_f where s(t_f) = 0 = -16t_f² +1351 ⇒ 16t_f² = 1351 ⇒ t_f = 9.189

e) v(9.189) = -294.048 ft/s