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3 years ago

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A 3.9 kg ball traveling towards a soccer player at a velocity of -3.5 m/s rebounds off the soccer player's foot at a velocity of

+15.9 m/s. If the Force of contact between the ball and the player's foot was 25.9 N. For how long was the force applied?

1 answer:

Tresset [83]3 years ago

4 0

Answer: 2.92 s

Explanation:

Given

Mass of ball is $m=3.9\ kg$

The initial velocity of the ball is $u=-3.5\ m/s$

Velocity after the rebound is $v=15.9\ m/s$

Force during the contact is $F=25.9\ N$

We know, change in momentum is Impulse

$\Rightarrow F\cdot \Delta t=m(\Delta v)$

$\Rightarrow 25.9\cdot \Delta t=3.9(15.9-(-3.5))\\\\\Rightarrow \Delta t=\dfrac{3.9\times 19.4}{25.9}=2.92\ s$

Thus, the force is applied for 2.92 s

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So, the time needed before you hear the splash is approximately <u>2.06 s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses two principles, namely the time for an object to fall freely and the time for sound to propagate through air. When moving in free fall, the time required can be calculated by the following equation:

$\sf{h = \frac{1}{2} \cdot g \cdot t^2}$

$\sf{\frac{2 \cdot h}{g} = t^2}$

$\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}$

With the following condition :

t = interval of the time (s)
h = height or any other displacement at vertical line (m)
g = acceleration of the gravity (m/s²)

Meanwhile, for sound propagation (without sound reflection), time propagates is the same as the quotient of distance by time. Or it can be formulated by :

$\boxed{\sf{\bold{t = \frac{s}{v}}}}$

With the following condition :

t = interval of the time (s)
s = shift or displacement (m)
v = velocity (m/s)

<h3>Problem Solving</h3>

We know that :

h = height or any other displacement at vertical line = 19.6 m
g = acceleration of the gravity = 9.8 m/s²
v = velocity = 343 m/s

What was asked :

$\sf{\sum t}$ = ... s

Step by step :

Find the time when the object falls freely until it hits the water. Save value as $\sf{\bold{t_1}}$

$\sf{t_1 = \sqrt{\frac{2 \cdot h}{g}}}$

$\sf{t_1 = \sqrt{\frac{2 \cdot \cancel{19.6} \:_2}{\cancel{9.8}}}}$

$\sf{t_1 = \sqrt{4}}$

$\sf{\bold{t_1 = 2 \: s}}$

Find the time when the sound propagate through air. Save value as $\sf{\bold{t_2}}$

$\sf{t_2 = \frac{h}{v}}$

$\sf{t_2 = \frac{19.6}{343}}$

$\sf{\bold{t_2 \approx 0.06 \: s}}$

Find the total time $\sf{\bold{\sum t}}$

$\sf{\sum t = t_1 + t_2}$

$\sf{\sum t \approx 2 + 0.06}$

$\boxed{\sf{\sum t \approx 2.06}}$

<h3>Conclusion</h3>

So, the time needed before you hear the splash is approximately 2.06 s.

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