Question

Base jumping at an elevation of 7569 feet the acceleration of gravity is 32.174 feet/second squared and my mass is 245 pounds. What is the velocity immediately after impact in miles per hour?

Answer 1

you can use the kinematical equation
\[v^{2}=u^{2}+2gs\]
to find the velocity just before the impact, where, v is the velocity just before the impact, u is the initial velocity which is 0 here, g is the acceleration due to gravity and s is the distance or displacement and here, it is 7569ft
the velocity just after impact seems to be something weired as you may die by hitting the ground and the velocity after impact will be 0 or your paraschute may save you from hitting the ground and you veocity befiore and after impact will be 0 again.
but if this is a question of mommentum conservation, momentum is not conserved when an impulse acts upon the body and this is a case where an impulse is assumed to be acted upon you
soaccording to me, we may need more information to solve this question

Answer 2

you can use conservation of energy.
\[\Delta KE+\Delta PE=0\]
\[\frac{ 1 }{ 2 }m \Delta (v^2)+mg \Delta h=0\]

Answer 3

Although it seems that @rajat97 is correct in that velocity just after impact is 0