At time t = 0 s a stone is thrwon from ground level straight up with a speed of

Question

At time t = 0 s a stone is thrwon from ground level straight up with a speed of 20 m / s.Ignore air friction. You may take g = 10 m / s^2.

(a) At what time will the stone reach its highest point, and how high is it above the ground at that time? The answer is 2.0s and 20m.

(b)

A second stone is thrown straight up 2 seconds after the first . How many meters above the ground is the first stone at that moment? The answer is 20m.

(c) At what speed should the second stone be thrown from the ground if it is to hit the first stone 1 second stone is thrown?

Please show how to do these problems...

Explanation / Answer

Part A)

Apply vf = vo +at

0 = (20) + (-10)(t)

t = 2 sec

d = vot + .5at^2

d = (20)(2) + (.5)(-10)(4)

d = 20 m

Part B)

Since it only takes 2 seconds for the first stone to be at the top of its path, that is where is is when you thrown the second stone. We found in part A, that the stone is at 20 m

Part C)
The first stone will be on its way down, and we can find out where it will be

d = (20)(3) + (.5)(-10)(9)

d = 15 m above the ground

Now we need to apply

d = vot + .5at^2

15 = (vo)(1) + (.5)(-10)(1)

vo = 20 m/s

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