A silver dollar is dropped from the top of a building that is 1350 feet tall. Us

Question

A silver dollar is dropped from the top of a building that is 1350 feet tall. Use the position function below for free-falling objects. s(t) = -16t^2 + V_0 t + S_0 (a) Determine the position and velocity functions for the coin. s(t) = v(t) = (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. () t = s (e) Find the velocity of the coin at impact. () ft/s

Explanation / Answer

Solution :

(a) Since the coin is falling freely, thus initial velocity v0 = 0 and s0 = 1350 feet.

The position function for the coin

s(t) = -16t^2 + v0t + s0

s(t) = -16t^2 + (0) t + 1350

Thus position function is:

s(t) = -16t^2 + 1350

And

To get velocity function, differentiate the position function with respect to t, to get

v(t) = ds(t)/dt

v(t) = (d/dt)(-16t^2 + 1350)

= -32t + 0

v(t) = -32t.

(b) For the average velocity, the distance at t = 3 sec.

s(3) = -16(3)^2 + 1350

Distance at t = 4 sec.

s(4) = -16(4)^2 +1350

Average velocity = total distance during the interval [3, 4] / time interval

= s(4) - s(3) / 4 - 3

= -16(4)^2 +1350 - (-16(3)^2 +1350) /1

= - 256 +144

= - 112

= 112 feet/sec.

(c) Instantaneues velocity at t = 3 sec,

v(t) = -32(3)

v(3) = 96 feet/sec.

Instantaneues velocity at t = 4 sec,

v(t) = -32(4)

v(4) = 128 feet/sec.

(d) When the coin reach the ground, then s(t) = 0.

Thus 0 = -16t^2 + 1350

t^2 = 1350/16

Taking square root of both side

t = 36.74 / 4

Thus t = 9.185 sec.

(e) At impact t = 9.185 sec

v(t) = -32(9.185)

= 293.920 feet/sec. Ans.

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