Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

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