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ddd [48]
3 years ago
8

Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter th

an if you land on water. Using the impulse-momentum theorem as a guide, determine which one of the following statements is correct.
(a) In bringing you to a halt, the sand exerts a greater impulse on you than does the water.

(b) In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force.

(c) In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a smaller average force.
Physics
1 answer:
denis-greek [22]3 years ago
7 0

Answer:

In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force

Explanation:

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Part of the question is missing. Here it is:

<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table.     The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity  just before it hit the floor? </em>

Answer:

\theta=-75.7^{\circ}

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

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We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

s=ut+\frac{1}{2}at^2

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

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a=g=9.8 m/s^2 is the acceleration of gravity

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Solving for t,

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s

Now we can find the final vertical velocity of the ball, using:

v_y=u+at

And susbtituting t = 0.52 s, we find

v_y = 0 +(9.8)(0.52)=5.1 m/s

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

v_x = \frac{0.7}{0.52}=1.3 m/s

So now we can find the direction of the ball's velocity using:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

\theta=-75.7^{\circ}

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I think is E
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