nainabhandari85
Answered

Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

A student on a tower 49 m height drops a stone. One second later he throws a second stone after the first. They both hit the ground at the same time, with what speed did he throw the second stone. [Ans: 10.1m/s]​

Sagot :

Lanuel

By applying the second equation of motion, the speed at which he threw the second stone is equal to 12.10 m/s.

How to determine the speed?

First of all, we would calculate the time taken by the first stone to reach a height of 49 meters by applying the second equation of motion as follows:

S = ut + ½gt²

49 = 0(t) + ½ × 9.8 × t²

49 = 4.9t²

t² = 49/4.9

t = √10

t = 3.16 seconds.

Now, we can determine the speed at which he threw the second stone:

Note: Time = 3.16 - 1 = 2.16 seconds.

S = ut + ½gt²

49 = u(2.16) + ½ × 9.8 × 2.16²

49 = 2.16u + 22.86

2.16u = 49 - 22.86

u = 26.14/2.16

u = 12.10 m/s.

Read more on initial speed here: https://brainly.com/question/19365526

#SPJ1

Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.