nicolliasimao
Answered

Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

A stone is thrown vertically upward at a speed of 41.50 m/s at time t=0. A second stone is thrown upward with the same speed 3.020 seconds later. At what time are the two stones at the same height?

At what height do the two stones pass each other?

What is the upward speed of the second stone as they pass each other?

Sagot :

Answer:

t1 = t2 + 3.02         V = 41.5

V t1 - 1/2 g t1^2 = V t2 - 1/2 g t2^2

Both stones reach the same height after the specified times

V (t1 - t2) = g/2 (t1^2 - t2^2) = g/2 (t1 - t2) (t1 + t2)

2 V / g = t1 + t2 = 2t1 + 3.02

t1 = V / g - 1.51 = 41.5 / 9.8 -1.51 = 2.72 s

t2 = t1 + 3.02 = 5.74 sec

Check:

41.5 * 2.72 - 4.9 * 2.72^2 = 76.6 m

41.5 * 5.74 - 4.9 * 5.74^2 = 76.8 m

Speed of second stone = 41.5 - 9.8 * 2.72 = 14.8 m/s

Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.