.) Given: stone is released from an elevator moving up at a speed of 12 f/s and

Question

.) Given: stone is released from an elevator moving up at a speed of 12 f/s and reaches the bottom of the elevator shaft n 2.5 s. A FBD of the stone with the origin of the x-axis located at the bottom of the elevator shaft is shown below. The stone does not bounce. Please, use the att Particles. Find: (a) State the case based on the form of the acceleration from the attached handout. (b) State the final height of the stone in ft (x). (c) State the final time when the stone strikes the bottom of the shaft in s (t). (d) State the initial speed of the stone in ftls (vo. (e) State the value of the acceleration in ft/s with the proper sign (a). (f) Calculate the initial height of the stone in ft (xo (g) Calculate the final speed of the stone in fts with the proper sign (v). ached handout for the Rectilinear Motion of s S TONE h-2.5s

Explanation / Answer

Let upward direction be positive for velocity and acceleration.

Distance calculation is referenced from the bottom of the elevator shaft which is the origin.

Gravitational acceleration acts on the stone.

Gravitational acceleration, a = -32.185 ft/s2 (Negative sign because it acts in downward direction)

Due to initial velocity of the lift, the stone will first go up till its velocity becomes zero. After this, the stone will start falling downwards due to gravity.

So, initial condition is when the stone is released.

Intermediate condition is when velocity of the stone reaches zero after reaching a certain height.

Final condition is just before touching the bottom of the elevator shaft.

When the stone travels up,

Initial condition (t=0)

Let the height from origin from where the stone is released be x0 ft.

Initial velocity (v0) = 12 ft/s

Intermediate condition

Intermediate velocity (vi) = 0 ft/s

Let the height travelled be h ft and the time taken be t1 sec.

vi = v0 + a t1

=> 0 = 12 - 32.185 t1

=> t1 = 0.373 sec

h = v0 t + 0.5 a t2

=> h = 12 x 0.373 - 0.5 x 32.185 x 0.3732 = 2.237 ft

When the stone falls down,

Intermediate condition

Intermediate velocity (vi) = 0 ft/s

Final condition

Let final velocity be v ft/s.

Total distance travelled from intermediate position (d) = 2.237 + x0

Given that total time taken from initial condition to final condition (t) = 2.5 sec.

So, time taken for the stone to fall from intermediate position to final position (t2) = 2.5 - 0.373 = 2.127 sec

v = vi + a t2

=> v = 0 - 32.185 x 2.127 = -68.463 ft/s (Negative sign because velocity is in downward direction)

Final position (x) = 0 ft (from origin)

Now, d = vi t2 + 0.5 a t22

=> -(2.237 + x0 ) = 0 - 0.5 x 32.185 x 2.1272 (Negative sign is used in LHS because the stone travels in downward direction)

=> x0 = 70.578 ft

(a) Gravitational acceleration acts on the stone.

(b) Final height of the stone (x) from origin = 0 ft

(c) Final time when the stone strikes the bottom of the shaft (t) = 2.5 sec

(d) Initial velocity of the stone (v0) = 12 ft/s (in upper direction)

(e) Acceleration value (a) = -32.185 ft/s2 (acts in downward direction)

(f) Initial height of the stone (x0) = 70.578 ft

(g) Final velocity of the stone (v) = -68.463 ft/s (acts in downward direction)

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