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sattari [20]
3 years ago
11

A marble, rolling with speed of 20m/sec rolls off the edge of the table that is 180m high (g=10m/sec2), find time taken to drop

to the ground.
a. 10sec
b. 8sec
c. 6sec
d. 12sec​
Physics
2 answers:
natali 33 [55]3 years ago
6 0

Answer:

<em>Choice: c. 6sec</em>

Explanation:

<u>Horizontal Launch </u>

When an object is thrown horizontally with a speed (v) from a height (h), it describes a curved path ruled by gravity until it finally hits the ground.

The horizontal component of the velocity is always constant because no acceleration exists in that direction, thus:

v_x=v

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

v_y=g.t

Where g=10 m/s^2

To calculate the time the object takes to hit the ground, we use the same formula as for free-fall, since the time does not depend on the initial speed:

\displaystyle t=\sqrt{\frac{2h}{g}}

The marble rolls the edge of the table at a height of h=180 m, thus:

\displaystyle t=\sqrt{\frac{2*180}{10}}

\displaystyle t=\sqrt{36}

t = 6 sec

Choice: c. 6sec

Monica [59]3 years ago
5 0

Answer:

The answer is:

C) 6sec

Hope it helps:D

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Question 30
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Answer: 0.69\°

Explanation:

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3 years ago
A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.
SVETLANKA909090 [29]

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:

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#LearnwithBrainly

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Geothermal, Wind ,Solar, and Water turbine.

I hope I helped.
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