Question

Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 10 m into 33 cm. of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 64 kg, find the magnitude of the impulse on him from the water.

Answer 1

Answer:

Impulse: $J=896Ns$

Explanation:

The impulse can be calculated with the average force applied by the water to Henri:

Impulse: $J=F_{av}*t_{stop}$

as from newton´s second law:

$F_{av}=m*a_{av}$

(1)$J=m*a_{av}*t_{stop}$

The average acceleration he experiments from the water can be calculated as:

$a_{av}=(V_{2}-V_{1})/t_{stop}$

As he stops at the bottom $V_{2}=0$, replacing in (1):

(2)$J=m*a_{av}*t_{stop}=m*(V_{1}/t_{stop})*t_{stop}=m*V_{1}$

To calculate the velocity when he reaches the water:

(3)$V_{1}=V_{0}+g*t$

As he starts from rest $V_{0}=0$, we need to use the kinematic equation for position to find the time it takes to reach the water:

$X=x_{0}+V_{0}*t+1/2g*t^{2}$

Assuming it starts from $x_{0}=0$ to reach X=10m:

$10m=1/2g*t^{2}$

solving for t:

$t=\sqrt{\frac{2*10m}{g} } =\sqrt{\frac{20m}{9,8m/s^{2} } } =1,43s$

replacing this time in (3):

$V_{1}=9,8m/s^{2}*1,43s=14m/s$

replacing this in (2):

$J=m*V_{1}=64kg*14m/s=896Ns$

Answer 2

The magnitude of the impulse is -896Ns.

How to calculate the magnitude of the impulse?

From the information given, the following can be deduced:

Height = 10m

Mass = 64kg

The kinetic energy is calculated as 1/2mv² while potential energy is calculated as mgh. From the law of conversation:

1/2mv² = mgh

v = ✓2gh

= ✓2 × ✓9.8 × ✓10

= 14m/s

The change in velocity will be;

= 64(0 - 14)

= 64 × -14

= -896Ns

In conclusion, the magnitude of the impulse is -896Ns.

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