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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 2 / Problem 85P

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 up,

Chapter 2, Problem 85P

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QUESTION:

Problem 85P

Look Out Below. Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it 2.20 m above the ground. 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:

Problem 85P

Look Out Below. Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it 2.20 m above the ground. 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,

$t=\sqrt{0.0366}=0.191\ s$

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

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