ZuoTi.Pro
Question

Chapter 2 roblem 2.73 10 of 12> Constants Sam heaves a shot with weight 16-lb straight upward, giving it a constant upward acceleration from rest of 34.7 m/s for a height 69.0 cm. He releases it at height 2.12 m above the ground. You may ignore air resistance. Correct Part B How high above the ground does it go? h4.56 m Correct Part 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, a distance 1.85 m above the ground 4.56-h t- 1-4.905 Submit Previous Answers Request Answer X Incorrect: Try Again: 3 attempts remaining
science   physics   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

ven ta SP bren erf C5) 69.ch: o.cqm acce le ra fon 3-) m/s2 From eava ton of roton . Ч.560692 m Tirre 씨ren

0 0
Add a comment
Know the answer?
Add Answer to:
Chapter 2 roblem 2.73 10 of 12> Constants Sam heaves a shot with weight 16-lb straight...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions

  • Part c Chapter 2 Problem 2.73 Part A Constants Sam heaves a shct with weight 16-lb...

    Part c Chapter 2 Problem 2.73 Part A Constants Sam heaves a shct with weight 16-lb straight upward, giving it a constant upward acceleration from rest of 347 m/st for a height 69.0 cm. He releases it at height 2.12 m above the ground. You may ignore air resistance. What is the speed of the shot when he releases it? 6.92 m/s Previous Answers Correct Part B How high above the ground does it go? h 4.56 m Correct Part...

  • Sam heaves a bowling ball with weight 16-lb straight upward, giving it a constant upward acceleration...

    Sam heaves a bowling ball with weight 16-lb straight upward, giving it a constant upward acceleration from rest of 44.3 m/s2 for a height 60.0 cm . He releases it at height 2.12 m above the ground. You may ignore air resistance. Answered A and B but stuck on C. Part A Part complete What is the speed of the bowling ball when he releases it? v=7.9m/s Part B Part complete How high above the ground does it go? h=4.83m...

  • CensantsPartA Sam heaves a shot with weght 16-1b upnind giving it a constant upwand acoeteraion trom...

    CensantsPartA Sam heaves a shot with weght 16-1b upnind giving it a constant upwand acoeteraion trom rest of 347m/ for a height 60 E. He releases in at heght 2.12 m above he ground. You may ignore ar resistance Wa·isthe weed of Po shot when he Mleases." m/s Part How righ above the gound does itgo? Part C How much Sme does he have to get out of its way before it etums to tPhe height of the top ofhis...

  • part A,B and C please Part A Constants What is the speed of the shot when...

    part A,B and C please Part A Constants What is the speed of the shot when he releases it? Sam heaves a shot with weight 16-lb straight upward, giving ita constant upward acceleration from rest of 42.8 m/s for a height 68.0 cm. He releases it at height 2.25 m above the ground. You may ignore air resistance. u= m/s Submit Part B How high above the ground does it go? Submit PartC How much time does he have to...

  • Exercise 3.18: Part C and D Constants A shot putter releases the shot some distance above...

    Exercise 3.18: Part C and D Constants A shot putter releases the shot some distance above the level ground with a velocity of 120 m/s 510 'above the horizontal The shot hits the ground 2 08 s later You can iqnore air resistance Part A What is the x-component of the shof's acceleration while in flight? Previous Answers Correct Part B What is the y-component of the shot's acceleration while in fight? ay -980 m/s? Previous Answers v Correct e...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT