[SOLVED] Physics problem w/ guns

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September 11th 2008, 09:11 PM
john doe

[SOLVED] Physics problem w/ guns

Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impact an initial speed of 30.0 m/s to each pellet. Gun A is fired straight upward, with the pellet going up 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?

when i look at this i saw that they only gave us 2 pieces of information
there's gravity and initial velocity i believe
September 12th 2008, 01:10 AM
ticbol

Quote:

Originally Posted by john doe

Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impact an initial speed of 30.0 m/s to each pellet. Gun A is fired straight upward, with the pellet going up 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?

when i look at this i saw that they only gave us 2 pieces of information
there's gravity and initial velocity i believe

Yes, the height of the cliff from the ground below is not that important in this problem.

This problem is more on your knowlefge of the parabolic flight of the bullet where air resistance is ignored.

The bullet fierd straight up will reach its max then it will fall down. At the maximum height, the vertical velocity is zero. Then at the same level where it was fired, the vertical velocity is the same as the initial velocity but in opposite direction.
Here it will be -30m/s.
That is the same as the initial velocity of the second bullet that is fired straight down.
That means the second bullet will be ahead of the first bullet by the amount of time that the first bullet spent from cliff level to max height to cliff level again.

At the max height, the vertical velocity of the first bullet is zero,
v = Vo -gt
0 = 30 -gt
t = 30/g seconds.

When the bullet falls to the level it was fired, it takes the same t = 30/g seconds again.
So it takes the first bullet 2(30/g) seconds to be in the same level where the second bullet was fired.

Therefore, whatever is the height of the cliff from the ground below, the first bullet will always be 60/g seconds late than the second bullet.

Hence, if your g = 10 m/sec/sec, then the first bullet will be 60/10 = 6 seconds late in hitting the ground.
If your g = 9.8 m/sec/sec, it would be 60/9.8 = 6.12245 seconds late.
September 12th 2008, 08:30 AM
Soroban

Hello, john doe!

Quote:

Two identical pellet guns are fired simultaneously from the edge of a cliff.
These guns have an initial speed of 30.0 m/s.
Gun A is fired straight upward, with the pellet going up 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?

The formula is: . $h \;=\;h_o + v_ot - 4.9t^2$ . where: . $\begin{Bmatrix} h &=& \text{height of pellet} \\ h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \end{Bmatrix}$

. . $\begin{array}{cccc}\text{Gun A:} & h &= & h_o + 30t - 4.9t^2 \\ \\[-3mm]<br /> \text{Gun B:} & h &=& h_o -30t - 4.9t^2\end{array}$

When does the pellet from Gun A hit the ground?

. . $4.9t^2 - 30t - h_o \:=\:0 \quad\Rightarrow\quad t_A \;=\;\frac{30 \pm\sqrt{900 + 19.6h_o}}{9.8}\:$ seconds

When does the pellet from Gun B hit the ground?

. . $4.9t^2 + 30t - h_o \:=\:0 \quad\Rightarrow\quad t_B \;=\;\frac{-30 \pm\sqrt{900 +19.6h_o}}{9.8}\:$ seconds

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

. . . . . . . . . . . . $= \;\frac{60}{9.8} \;=\;\frac{300}{49} \;\approx\; 6.12\text{ seconds}$
September 12th 2008, 10:38 AM
topsquark

Quote:

Originally Posted by Soroban

Hello, john doe!

The formula is: . $h \;=\;h_o + v_ot - 4.9t^2$ . where: . $\begin{Bmatrix} h &=& \text{height of pellet} \\ h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \end{Bmatrix}$

. . $\begin{array}{cccc}\text{Gun A:} & h &= & h_o + 30t - 4.9t^2 \\ \\[-3mm]<br /> \text{Gun B:} & h &=& h_o -30t - 4.9t^2\end{array}$

When does the pellet from Gun A hit the ground?

. . $4.9t^2 - 30t - h_o \:=\:0 \quad\Rightarrow\quad t_A \;=\;\frac{30 \pm\sqrt{900 + 19.6h_o}}{9.8}\:$ seconds

When does the pellet from Gun B hit the ground?

. . $4.9t^2 + 30t - h_o \:=\:0 \quad\Rightarrow\quad t_B \;=\;\frac{-30 \pm\sqrt{900 +19.6h_o}}{9.8}\:$ seconds

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

. . . . . . . . . . . . $= \;\frac{60}{9.8} \;=\;\frac{300}{49} \;\approx\; 6.12\text{ seconds}$

Nicely done, though I'm going to get picky about one thing: Since $30 < \sqrt{900 + 19.6h_0}$ for any $h_0 > 0$ the negative sign before the square root in the $t_A - t_B$ equation is invalid. So it should read
$t_A - t_B \;=\;\left(\frac{30 + \sqrt{900+19.6h_o}}{9.8}\right) - \left(\frac{-30 + \sqrt{900 + 19.6h_o}}{9.8} \right)$
which, of course, leads to the same result Soroban already gave, so this is only a small matter. I include it only for the sake of completeness.

-Dan