# Math Help - What is the V at the ground

1. ## What is the V at the ground

I have Gun A with Vo of 30m/s shot upward into the air and Gun B shot towards the ground at the same Vo.
Now I know a for A is -9.8 and a for B is 9.8.
As Vo is 30m/s when it hits the ground is it 0 m/s. That is what I am not sure of.

But I did calc time on Gun A with t=V-Vo/a and get 3.06S. which is from the cliff to the top of the curve before it falls.

How long after Gun B's pellet hits the ground does Gun A's pellet hit the ground.
I know they hit the ground at the same time. as the Vo on the way down from Gun B's position is 30m.s.

The answer is 6.12s, and if I dbl the 3.06 s above I get that, but that does not make sense to me. Can someone explain.
Tks

Sorry, most of what you said makes no sense.

I have Gun A with Vo of 30m/s shot upward into the air
and Gun B shot towards the ground at the same Vo. . From what height?

Now I know a for A is -9.8 and a for B is 9.8.
No ... gravitational acceleration does not reverse
just because the projectile is shot downward.

As Vo is 30m/s, when it hits the ground, is it 0 m/s ?
Certainly not!
It is 0 m/s after it hits the ground.

If you don't believe it, let me drop an anvil on your head.

But I did calc time on Gun A with t=V-Vo/a and get 3.06S. which is from the cliff
to the top of the curve before it falls. . What cliff?

How long after Gun B's pellet hits the ground does Gun A's pellet hit the ground?
I know they hit the ground at the same time.
If they hit the ground together, the question makes no sense.

3. Sorry, I guess it does not make sense. No height is given. The 2 guns are fired from a cliff. One up into the air and the other one straight down to the ground.

They are both shot at the same time the two guns off the cliff. and they both impart an intial speed of 30 m/s.
The absense of air resistance, how long after pellet B which is the one going straight to the ground below the cliff, does the first pellet which was shot into the air hit the ground??
Does it make sense now.?? Sorry, about that.

Now I get it . . .

Two guns are fired from the top of a cliff.
One (gun $\,A$) up into the air and the other (gun $\,B$) straight down to the ground.

They are both shot at the same time with an intial speed of 30 m/s.
The absense of air resistance, how long after pellet $\,B$ hits the ground
does pellet $\,A$ hit the ground?

Gun $\,A$ has the height function: . $h_A \;=\;h_o + 30t - 4.9t^2$

Gun $\,B$ has the height function: . $h_B \;=\;h_o - 30t - 4.9t^2$

Pellet $\,A$ hits the ground when $h_A \,=\,0.$

. . $h_o + 30t - 4.9t^2 \:=\:0 \quad\Rightarrow\quad 4.9t^2 - 30t - h_o \:=\:0$

. . Quadratic Formula: . $t \;=\;\dfrac{30 \pm\sqrt{900 + 19.6h_o}}{9.8}$

. . Pellet $\,A$ hits the ground in: . $t_A \:=\:\dfrac{30+\sqrt{900+19.6h_o}}{9.8}$ seconds.

Pellet $\,B$ hits the ground when $h_B \,=\,0.$

. . $h_o - 30t - 4.9t^2 \:=\:0 \quad\Rightarrow\quad 4.9t^2 + 30t - h_i \:=\:0$

. . Quadratic Formula: . $t \;=\;\dfrac{-30 \pm\sqrt{900 + 19.6h_o}}{9.8}$

. . Pellet $\,B$ hits the ground in: . $t_B \;=\;\dfrac{-30 + \sqrt{900 + 19.6h_o}}{9.8}$ seconds.

The difference of the times is:

. . $t_A - t_B \;=\; \left(\dfrac{30 + \sqrt{900+19.6h_o}}{9.8}\right) - \left(\dfrac{-30 + \sqrt{900+19.6h_o}}{9.8}\right)$

. . . . . . . . $=\; \dfrac{60}{9.8} \;=\;\dfrac{300}{49} \;=\;6\frac{6}{49}\text{ seconds.}$