Two identical pellet guns are fired simultaneously from the edge of a cliff. The

Question

Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 38 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? (in sec)

I started by labeling knowns and unknowns:
VoA= 38 m/s
accelA= -9.8 m/s^2 (gravity)
timeA=?

VoB = -38 m/s
accelB=-9.8 m/s^2
timeB=?

Then wrote my equations of motion:
VA = VoA + accelA(timeA)
VB = VoB + accelB(timeB)

Is it safe to assume both VA and VB (velocities of A and B respectively) are = 0?
(Since both pebbles hit the ground at this instant, their velocity is 0?)

Either way, I'm stuck trying to solve... Someone help please!

Explanation / Answer

Let:
u be the firing speed of the gun,
h be the height reached by pellet A above the clifftop,
t be the required time difference,
g be the acceleration due to gravity.

When pellet A arrives back at the starting level, it has velocity u downwards.
This is identical to the initial velocity of pellet B.
The descent from the cliff therefore takes the same time for each pellet.

The additional time required by pellet A is the time taken to ascend and return to the starting point.
0 = ut - gt^2 / 2
t(u - gt / 2) = 0

The non-zero root is:
t = 2u / g
= 2 * 38 / 9.81
= 7.747 sec.

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