Find Time for 2 Pellet Guns Fired with Initial Speed of 30 m/s

[afcwestwarrior]

Homework Statement

2 Identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 30.0 m/s to each pellet. Gun A is fired straight upward, with the pellet going up the cliff and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward. In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground.
a= 9.8 m/s^2
Vo=30.0 m/s
t=?

Homework Equations

y=Vot + 1/2 at^2
or t=Vo*2/a
v=Vo + at

The Attempt at a Solution

Here is what I did. 30.0 m/s *2/9.8 m/s^2= 6.12 s. The answer is right, but did I do the problem right.

 

[learningphysics]

How did you come up with that?

 

[bel]

Well, $$ms^{-1}$$ times $$ms^{-2}$$ would not give an answer in seconds. Since both would fall from the cliff with an initial velocity of $$30 ms^{-1}$$ and with the same acceleration, the time difference would be the time taken for pellet A to go up and come back down to the level of the cliff.

 

[learningphysics]

Well, $$ms^{-1}$$ times $$ms^{-2}$$ would not give an answer in seconds. Since both would fall from the cliff with an initial velocity of $$30 ms^{-1}$$ and with the same acceleration, the time difference would be the time taken for pellet A to go up and come back down to the level of the cliff.

No it's the right answer... the units come out right. it's ms^-1 divided by ms^-2.

 

[afcwestwarrior]

so did i do it right

 

[olgranpappy]

so did i do it right

You have not told us what you did to arrive at the expression (in particular the lefthand side of the expression):

"30.0 m/s *2/9.8 m/s^2= 6.12 s"

Thus we can not yet tell you if what you did was right.