Ok, this seems like a simple enough problem, however I have trouble wrapping my mind around physics concepts and trying to set up these problems. Any help will be greatly appreciated. Here is how the problem reads:
Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart and 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?
As much detail as you can include will help me greatly. I have quite a few of these problems to work and hope I can use this as an example. Thanks again to anyone who tries this.
2007-09-12 15:16:34 · 1 answers · asked by doubtful 2 in Science & Mathematics ➔ Physics
The key thing to notice here is that after pellet A goes up and then comes back down to your level, at that instant it will be travelling downward at 30 m/s; which means it will be exactly like a "delayed" version of pellet B.
The amount of the delay will be exactly the amount of time it takes for pellet A to go up and then back down to your level.
From the formula g = Δv/t, we get:
t = Δv/g
= (30 m/s – (-30 m/s))/(9.8 m/s²)
= 6.1 sec.
2007-09-12 15:35:43 · answer #1 · answered by RickB 7 · 1⤊ 0⤋