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Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford
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Look Out Below. Sam heaves a 16-lb shot straight upward,

Chapter 2, Problem 2.85

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QUESTION:

Look Out Below. Sam heaves a 16-lb shot straight upward, giving it a constant upward acceleration from rest of \(35.0\mathrm{\ m}/\mathrm{s}^2\) for 64.0 cm. He releases it 2.20 m above the ground. You may ignore air resistance.

(a) What is the speed of the shot when Sam releases it?

(b) How high above the ground does it go?

(c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?

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QUESTION:

Look Out Below. Sam heaves a 16-lb shot straight upward, giving it a constant upward acceleration from rest of \(35.0\mathrm{\ m}/\mathrm{s}^2\) for 64.0 cm. He releases it 2.20 m above the ground. You may ignore air resistance.

(a) What is the speed of the shot when Sam releases it?

(b) How high above the ground does it go?

(c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?

ANSWER:

Step 1 of 5

Provided, the acceleration of the 16 lb shot as, .

The distance with which it travels for -  acceleration is, d = 64 cm = 0.64 m

Therefore, we can use,  relation in order to calculate the time.

Here,

We know that, initially, the shot was at rest. Hence, u = 0 m/s

Therefore,

Or,

So, the shot will take 0.191 s to accelerate up to  within 0.64 m distance.

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