Question

At the same moment from the top of a building 3.0 × 10 2 m tall, one rock is dropped and one is thrown downward with an initial velocity of 23 m/s. Both of them experience negligible air resistance. How much EARLIER does the thrown rock strike the ground?

Answer 1

multiply then divide?

Answer 2

multiply what and divide what

Answer 3

use the speed formula.

Answer 4

3.0 × 10 2

Answer 5

V = d/t i suppose you are required to find time. rearrange equation to find time.

Answer 6

how i did 3.0*10^2/23 but the answer is not correct

Answer 7

where u go

Answer 8

https://answers.yahoo.com/question/index?qid=20130609031821AAsLSHq

You need to use the equation:

s = ut + (!/2)at^2

you can figure out the time that the dropped ball to hit the ground quite easily because u = 0 m/s, therefore s = (1/2)at^2

t^2 =2s/a

t = sqrt(2s/a)

assuming the acceleration from gravity is 9.81 m/s^2

t = sqrt(2*300m/9.81 m/s*2)

t = 7.8 s

You can use the quadratic equation to find t for the thrown rock and then subtract them. Or you can calculate the final velocity of the thrown rock with v^2 = u^2 + 2as and then use v = u + at to find the time for the thrown rock.

v^2 = (10 m/a)^2 + 2 * 9.81 m/s^2 * 300 m

v = 76.7 m/s

v = u + at

t = (v - u)/s

t = (76.7 m/s - 10 m/s)/9.81 m/s^2

t = 6.8 sec

So the throw rock hit the ground one second earlier than the dropped rock.

Answer 9

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Answer 10

isn't u = 23m/s?

Answer 11

i know throw rock hit the ground earlier but there are a choices -1.4 s
-1.8 s
-2 s