when a 65 kg skydiver jumps from a plane, and her kinetic energy is 7.05*10 to the 5th power J how fast is she falling
Can you use KE = 1/2 mv^2
To calculate the speed at which the skydiver is falling, we need to use the equation for kinetic energy:
Kinetic energy (KE) = 1/2 * mass * velocity^2
In this case, we know the kinetic energy (7.05*10^5 J) and the mass of the skydiver (65 kg). We can rearrange the equation to solve for velocity:
velocity = sqrt(2 * kinetic energy / mass)
Substituting the given values into the equation, we can calculate the velocity:
velocity = sqrt(2 * 7.05*10^5 J / 65 kg)
Simplifying the equation, we get:
velocity = sqrt(2 * (7.05*10^5 J / 65 kg))
velocity = sqrt(2 * 1.0846*10^4 m^2/s^2)
velocity = sqrt(2 * 10846) m/s
Using a calculator, we find:
velocity = sqrt(21692) m/s
velocity ≈ 147.4 m/s
Therefore, the skydiver is falling at approximately 147.4 m/s.