a sky diver jumps out of a plane at a height of 5000 ft. if her parachute does not open until she reaches 1000 ft, what is her velocity at that point if air resisitance is disreguared?

To find the velocity of the skydiver when her parachute does not open, we can use the principles of motion and the concept of conservation of energy.

Let's assume the initial height from which the skydiver jumps is 5000 ft and the height at which her parachute opens is 1000 ft. We'll also disregard air resistance, which means there will be no external force acting on the skydiver after jumping.

The potential energy of an object is given by the formula: PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.

At the initial height of 5000 ft:
PE_initial = m * g * h_initial

At the height of 1000 ft, the potential energy will be converted entirely into kinetic energy, as air resistance is disregarded:
KE = (1/2) * m * v^2, where v is the velocity

Therefore, the potential energy at this point is:
PE_at_1000ft = KE

Setting the two equations equal to each other:
m * g * h_initial = (1/2) * m * v^2

The mass cancels out, leaving us with:
g * h_initial = (1/2) * v^2

Simplifying further:
v^2 = (2 * g * h_initial)

Substituting the values, let's convert the height from ft to m (since g is given in m/s²):
h_initial = 5000 ft * (0.3048 m/ft) = 1524 m

Now, we can calculate the velocity at 1000 ft:
v^2 = 2 * 9.8 m/s² * 1524 m

v^2 = 29815.2 m²/s²

Therefore, the velocity at the point of the parachute opening is approximately:
v ≈ √(29815.2 m²/s²) ≈ 172.7 m/s

V^2 = Vo^2 + 2g*d

V^2 = 0 + 64*(5000-1000) = 256,000
V = 506 Ft/s