2 identical pellet guns are fired at the same time from edge of cliff.they impart initial speed30m/s each.?

Question

gun A is fired straight upward with pellet going up and then falling back down eventually hitting ground beneath the cliff. Gun B is fired downwards. in absence of air resistance how long after pellet B hits the ground does pellet A hit the ground?

Answer 1

The difference in time is only from the point pellet A is shot in the air to the point where it falls to the level of the edge of the cliff. Once pellet A reaches the level of the edge of the cliff it will take it as much time to reach the ground as pellet B.

To solve for the time it takes pellet A to reach its highest point, use the formulas,

v^2-u^2=2as
and
s=ut+1/2at^2

Solve for s first.

Given: v=0
a=9.8m/s^2

v is the final speed at the highest point of the flight of pellet A. Substitute known values:

0-30^2=2*(-9.8)s
s= 900/19.6
s =45.91m

a is negative because pellet A is moving against gravity.

Now solve for t in the formula,

s=ut+1/2at^2

Substitute known values:

45.91=0*t+1/2*9.8t^2
45.91=4.9t^2
t^2=45.91/4.9
=9.36
t=3.05s

u is the initial speed at the highest point of the flight of pellet A as it descends towards the edge of the cliff. At that point the speed is equal to 0.

Multiply t by 2 to get the total time it takes pellet A to reach the highest point of its flight and fall down to the level of the edge of the cliff.

2*t=6.1s

Pellet A hits the ground 6.1 seconds after pellet B hits the ground.

Answer 2

gun A is fired up at 30 m /s rises to a height, stops, then falls back down to the starting position with the same velocity as starting velocity in opposite direction. proof is


up motion has height x Vf = 0, Vi = 30, a = -9.8
Vf^2 = Vi^2 + 2 a x
down motion has height x, Vi = 0, a = 9.8 m/s^2
x = Vot + 1/2 at^2

substituting x = -Vi^2/2a into second equation demonstrates that

since the pellet from A would then have the same initial velocity as the pellet from B after A completes it's up and down flight, then A arrives after B by the same amout of time it takes to go up and down = 2 x time up

so Vf = Vi + at (up)
0 = 30 m/s + (-9.8 m/s^2) x t
t (up) = (30 m/s) / (9.8 m/s^2) = 3.06 s

so t up and down = 6.12 sec.

pellet from A arrives 6.12 seconds after pellet from B

Answer 3

45 seconds if there were no birds or wind......just a guess