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A stone is dropped from a cliff 64 feet above the ground. Answer the following, using -32 ft/sec2 as the acceleration due to gravity. Show all work and submit to D2L.

a. Find functions that represent the acceleration, velocity, and position of the stone above the ground at time t.
b. How long does it take the stone to reach the ground?
c. With what velocity does the stone hit the ground?

User Adam Mrozek
by
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1 Answer

19 votes
19 votes

Answer:

(a) v = 32 t

h = 16 t^2

g = 32 ft/s^2

(b) 64 ft/s

Step-by-step explanation:

height, h = 64 feet

g = - 32 ft/s^2

(a) Let the time is t .

Let the velocity after time t is v.

Use first equation of motion

v = u + at

- v = 0 - 32 t

v = 32 t

Let the distance is h from the top.

Use second equation of motion


h = u t + 0.5 at^2 \\\\- h = 0 - 0.5* 32 * t^2\\\\h = 16 t^2

The acceleration is constant for entire motion.

(b) Let the velocity is v as it hits the ground. Use third equation of motion


v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 * 32* 64\\\\v = 64 feet/s

User Michael Rys
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