- #1
jenavira
- 10
- 0
This is driving me nuts.
A stone with weight w is thrown vertically upward into the air from ground level with initial speed v(i). If a constant force f due to air drag acts on the stone throughout its flight, (a) show that the maximum height reached by the stone is
h = v(i)^2 / 2g(1+f/w)
and (b) show that the stone's speed just before impact with the ground is
v = v(i) (w-f/w+f)^1/2
I've been working through these for ages and I'm just lost. I know I need the conservation of energy equation 1/2mv^2 = mgh and probably the kinetic motion equation 2a(delta x) = v(f)^2 - v(i)^2, but I'm just not sure what to do with them.
A stone with weight w is thrown vertically upward into the air from ground level with initial speed v(i). If a constant force f due to air drag acts on the stone throughout its flight, (a) show that the maximum height reached by the stone is
h = v(i)^2 / 2g(1+f/w)
and (b) show that the stone's speed just before impact with the ground is
v = v(i) (w-f/w+f)^1/2
I've been working through these for ages and I'm just lost. I know I need the conservation of energy equation 1/2mv^2 = mgh and probably the kinetic motion equation 2a(delta x) = v(f)^2 - v(i)^2, but I'm just not sure what to do with them.