Question

A person with mass mo drops from a diving platform into the pool ho meters below hat is the diver's gravitational potential energy when standing atop the platform? b. How much kinetic energy does the diver possess at impact? What is the diver's velocity vo at impact? d. What about a diver with mass m, -2mo? - what will their velocity v, be at impact? e. What if the diver with mass mo dives from a platform h-2ho meters - what will their velocity v2 be at impact? f. What height h,is necessary for the diver with mass mo to attain a velocity of 2vo

Answer 1

This question is purely based on the principle of conservation of energy.

a) Gravitational potential energy, U = mgh

= m₀gh₀

b) By conservation of energy, total energy should remains constant. So when it reaches at ground , tha man must loose all its potential energy, and all this potential energy should convert into kinetic energy.

So kinetic energy = potential energy = m₀gh₀

c) Let consider that , potential energy at the bottom is zero. So applying energy conservation principle

Kinetic energy_{at top} + potential energy_{energyat top} = kinetic energy _{at bottom} + potential energy_{at bottom}

0 + m₀gh₀ = 1/2m₀v₀² + 0

V₀² = 2gh₀

V₀ = √2gh₀

d) from the formula derived in c part of tge answer, velocity is independent of mass. So tha velocity V₁ at impact will be same as that of V_0.

e) if h₂= 2h₀

V₂ = √2gh₂ = √4gh₀

f) for V₃  to be equal to 2V₀

V₃ = 2V₀ = 2√2gh₀

= √2g4h₀

This means that h_{3 =} 4h₀