Question

A physics student throws a softball straight up into the air

A physics student throws a softball straight up into the air. The ball was in the air for a total of 6.16 s before it was caught at its original position.
(a) What was the initial velocity of the ball?
(b) How high did it rise?
0 0
Add a comment Improve this question Transcribed image text
Answer #1
Tr + Tf = 6.16 s,
Tf = Tr,
Tr + Tr = 6.16,
2Tr = 6.16,
Tr = 3.08 s. = Rise time or time to reach max. heightt.

a. Vf = Vo + gt,
Vo = Vf - gt,
Vo = 0 - (-9.8)*3.08 = 30.2 m/s.

b. h = Vo*t + 0.5g*t^2,
h = 30.2*3.08 - 4.9*(3.08)^2 = 46.53 m.
answered by: Keishia
Add a comment
Know the answer?
Add Answer to:
A physics student throws a softball straight up into the air
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

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
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