Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 44.7 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?

Bullet A hits the ground after 9 sec.

To find out how long after pellet B hits the ground pellet A hits the ground, we can analyze the motion of both pellets separately.

Let's start with pellet B. Since it is fired straight downward, its initial velocity is downward. The acceleration due to gravity is always directed downward, so we can consider it as a positive acceleration.

Using the kinematic equation for motion under constant acceleration, we can find the time it takes for pellet B to hit the ground. The equation we can use is:

d = vit + (1/2)at^2

where
d = displacement (distance traveled) of pellet B (which is equal to the height of the cliff)
vi = initial velocity of pellet B (44.7 m/s downward)
a = acceleration due to gravity (9.8 m/s^2)
t = time taken

Since pellet B is fired from the edge of the cliff, its displacement is equal to the height of the cliff. Therefore, we have:

-h = (44.7 m/s)(t) + (1/2)(9.8 m/s^2)(t^2)

where h represents the height of the cliff.

Now, let's move on to pellet A. It is fired straight upward, so its initial velocity is upward, and the acceleration due to gravity acts in the opposite direction (downward). We can still use the same kinematic equation, but with negative values for acceleration and displacement.

-h = (44.7 m/s)(t') - (1/2)(9.8 m/s^2)(t'^2)

where t' represents the time taken for pellet A to hit the ground.

To find the time difference between pellet A and pellet B hitting the ground, we need to find the values of t and t'.

One way to solve this problem is by solving the two equations simultaneously. Rearrange both equations to make h the subject:

(h - (1/2)(9.8 m/s^2)(t^2)) = (44.7 m/s)(t)
(h + (1/2)(9.8 m/s^2)(t'^2)) = (44.7 m/s)(t')

Setting these two expressions equal to each other, since they both equal the height of the cliff:

(h - (1/2)(9.8 m/s^2)(t^2)) = (h + (1/2)(9.8 m/s^2)(t'^2))

Simplify and rearrange the equation to solve for t':

(1/2)(9.8 m/s^2)(t^2) = (1/2)(9.8 m/s^2)(t'^2)

Cancel out common factors:

t^2 = t'^2

Take the square root of both sides:

t = t'

This equation tells us that the time taken for pellet A to hit the ground is equal to the time taken for pellet B to hit the ground. Therefore, the time difference between pellet A and pellet B hitting the ground is zero. Both pellets will hit the ground simultaneously.

Gun A:

V = Vo + g*Tr = 0.
Tr = -Vo/g = -44.7/-9.8 = 4.56 s. = Rise
time.

h = Vo*Tr + 0.5g*Tr^2 = 44.7*4.56 - 4.9*4.56^2 = 102 m.

Tf = Tr = 4.56 s. = Time to fall back
to top of cliff.

T = Tr+Tf = 4.56 + 4.56 = 9.12 s. = Time
to rise and fall back to top of cliff.

The time required for each pellet to
fall from top of cliff to ground is
equal. Therefore, pellet A hits ground
9.12 s after pellet B.