Question

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]​

Answer 1

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: brainly.com/question/19365526

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