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3 years ago

10

A mis-hit golf ball flies straight upward. It reaches a height of 31.0 meters before falling back down. How fast was it going wh

en it was hit? (Assume you can ignore drag)

1 answer:

Tema [17]3 years ago

5 0

At the top most position when ball reach to height 31 m

The speed of the ball must be ZERO

now we can use kinematics to find the final speed of the ball when it will hit back the ground

the equation of kinematics will be

$v_f^2 - v_i^2 = 2 ad$

$v_f^2 - 0^2 = 2(9.8)(31)$

$v_f^2 = 607.6$

$v_f = 24.65 m/s$

so it will hit the ground at speed 24.65 m/s

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A pulley system lifts a 1345 n weight a distance of 0.975m. Paul pulls the rope a distance of 3.90m, exerting a force of 375N. A

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A. IMA: 4

The Ideal Mechanical Advantage (IMA) is given by:

$IMA = \frac{d_i}{d_o}$

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$d_o$ is the output distance

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$MA=\frac{F_o}{F_i}$

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3 years ago

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2 years ago

Read 2 more answers

A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.

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A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

A)

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of $x_0 = 5.0 m$

- at t = 0, the initial velocity of the motorcyclist is $v_0 = 15.0 m$ east

- The acceleration of the motorcyclist is constant and it is $a=4.0 m/s^2$

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

$x(t)=x_0 + v_0t + \frac{1}{2}at^2$

where t is the time.

Substituting t = 2.0 s, we find the position:

$x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m$

B)

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

$v(t)=x'(t)=v_0 + at$

where:

$v_0=15.0 m/s$ is the initial velocity

$a=4.0 m/s^2$ is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

$t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s$

And therefore, the position at t = 2.5 s is:

$x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

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Answer:

A slow down and stop

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