A bullet is fired upward with a speed v0 from the edge of a cliff of height h (s
ID: 1706589 • Letter: A
Question
A bullet is fired upward with a speed v0 from the edge of a cliff of height h (see figure below). Ignore air drag. Assume the bullet is fired straight up in (a) and (b) and straight down in (c).
(a) What is the speed of the bullet when it passes by the cliff on its way down? (Use the following as necessary: v0.)
v =
(b) What is the speed of the bullet just before it strikes the ground? (Use the following as necessary: v0, g, and h.)
v =
(c) If the bullet is instead fired downward with the same initial speed v0, what is its speed just before it strikes the ground? (Use the following as necessary: v0, g, and h.)
v =
Explanation / Answer
Just plug it into kinematics,
a) v^2 = v(initial)^2 + 2a(x-x0)
since x-x0 is 0,
v^2=v(initial)^2
v=v0
b) Well, we already know that as it approaches the cliff, its speed is v0
so plug into another kinematic
v^2 = v0^2 + 2a(x-x0)
v^2 = v0^2 + (2)(-g)(-h)
v=(v0^2 + 2gh)
c) Well, we already know that v=v0, so if it is fired downward initially with v0, it is the same thing as b
v=(v0^2 + 2gh)
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