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Fundamentals Of Physics
Found in: Page 39
Fundamentals Of Physics

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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Short Answer

The Zero Gravity Research Facility at the NASA Glenn Research Center includes a 145 m drop tower. This is an evacuated vertical tower through which, among other possibilities, a 1 mdiameter sphere containing an experimental package can be dropped. (a) How long is the sphere in free fall? (b) What is its speed just as it reaches a catching device at the bottom of the tower? (c) When caught, the sphere experiences an average deceleration of 25 g as its speed is reduced to zero. Through what distance does it travel during the deceleration?

  1. The time spent by sphere in a free fall is5.44 s.

  2. The speed of the sphere as it reaches a catching device at the bottom of the tower is 53.3 m/s.

  3. The distance travelled by the sphere during deceleration is5.80 m.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given data

The height of the tower, y= 145 m

Step 2: Understanding the free fall acceleration

In free fall acceleration, objects accelerates vertically downward at constant rate. This constant acceleration is represented by g and it is known as free fall acceleration.

The kinematic equations under free falls is written as:

v= v0 +gt … (i)

y =v0t + 12 gt2 … (ii)

v2 =v02 +2 gy … (iii)

Here, v0 is the initial velocity, v is the final velocity, t is the time, g is the acceleration and y is the vertical displacement.

Step 3: (a) Determination of the time spent by sphere in a free fall

Using equation (ii), the time spent by sphere in a free fall is calculated as follows:

y =v0t + 12 gt2145 m =0 + 12(9.8 m/s2)t2 t2 = 2 × 1459.8 s2 t = 5.44 s

Therefore, the sphere is in free fall for5.44 s.

Step 4: (b) Determination of the speed of the sphere as it reaches a catching device

Using equation (iii), the speed of the sphere is calculated as follows:

v2 =v02 +2 gyv2 =0 + 2 (9.8 m/s2)(145 m)v = 2842 m/s = 53.3 m/s

Therefore, the speed of the sphere as it reaches a catching device at the bottom of the tower is 53.3 m/s.

Step 5: (c) Determination of the distance traveled during the deceleration

For decelerated motion, v = 0, v0 = 53.3 m/s and a = -25 g

Using equation (ii), the distance traveled during deceleration is calculated as follows:

y = v2- v022(-25 g) = (0 m/s)2-(53.3 m/s)22-(-25×9.8 m/s2) = 5.80 m

Therefore, sphere travels 5.80 m during deceleration.

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