If a paratroopers legs have enough strength to support five times her weight, and they can do so over a travel distance of 0.5 m as they flex, from what height can she free fall to a solid surface without injury

the required height is 2.5 m

Explanation:

given the given data in question

Lime to come in rest;

using S = ((U1 + U2)/2) t

⇒ 0.5 = ((√(2gh) + 0)/2) t

t = (0.5 × 2) / √(2gh)

now fat = impulse = change in momentum dp

F = 5mg = dp/dt = [(m√(2gh))/(0.5 × 2)] × √(2gh)

5mg = 2mgh/(0.5 × 2)

5 = h / 0.5

h = 5 × 0.5

h = 2.5 m

Therefore the required height is 2.5 m

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

trig 15.3 cos