A 70kg sky diver falls 5000m.
a) by how much does her gpe decrease during her fall? Ep=mxgxh = 70x10x5000 = 3,500,000J
b) By how much does her kinetic energy increase during the fall? KE GAINED = GPE LOST = 3,500,oooJ
c) What would her speed be just before she hit the ground?
d) Why is your answer to (c) wrong?
c) (1/2) m v^2 = answer to b
d) The reason they want is that the person would have reached terminal velocity where air drag up = weight down before falling 5000 meters.
The reason they forgot about is that the diver also has the constant horizontal velocity of the airplane the whole time. I guess the person jumped from a stationary hot air balloon :)
Thanks Damon!
To determine the speed just before the skydiver hits the ground, we can use the principle of conservation of energy. We know that the potential energy lost by the skydiver is equal to the kinetic energy gained.
b) By how much does her kinetic energy increase during the fall?
As stated, the kinetic energy gained is equal to the potential energy lost. So, the kinetic energy gained during the fall is also 3,500,000 Joules.
c) What would her speed be just before she hit the ground?
To calculate the speed, we can use the equation for kinetic energy:
Kinetic Energy (KE) = 1/2 * mass * velocity^2
Since we know the kinetic energy gained, we can rearrange the equation to solve for velocity:
velocity = sqrt((2 * KE_gained) / mass)
Plugging in the values:
velocity = sqrt((2 * 3,500,000 J) / 70 kg)
velocity = sqrt(100,000 J / kg)
velocity = 316 m/s
Therefore, the speed just before the skydiver hits the ground would be approximately 316 m/s.
d) Why is your answer to (c) wrong?
Without further explanation, it is difficult to determine what aspect of the answer could be considered incorrect. If there is a specific error in the calculations or an alternate interpretation of the question, please provide more context for clarification.