Question

Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 62 kg, find the magnitudes of the impulse on him from the water.

Answer 1

Answer:

823.46 kgm/s

Explanation:

At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.

So, mgh = 1/2mv²

From here, his velocity just as he reaches the surface of the water is

v = √2gh

h = 9 m and g = 9.8 m/s²

v = √(2 × 9 × 9.8) m/s

v = √176.4 m/s

v₁ = 13.28 m/s

So his velocity just as he reaches the surface of the water is 13.28 m/s.

Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.

So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.

His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,

J = (62 × 0 + 62 × 13.28) kgm/s = 0 +  823.46 kgm/s = 823.46 kgm/s

So the magnitude of the impulse J of the water on him is 823.46 kgm/s