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3 years ago

A ball is dropped from rest on a cliff what is the speed of the ball 5 seconds later

Physics

1 answer:

diamong [38]3 years ago

4 0

Answer:

The velocity of the ball after 5 seconds will be 49 m/s

Explanation:

v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

Initial velocity of the ball = 0 (As the ball is dropped from rest )

Acceleration due to gravity = 9.8 m/s

Time taken = 5 sec

As the acceleration due to gravity is constant in both the cases we can use the equations of motion in order to solve this question

Part I :- As we already know the values of u,g,ant t we can use the first equation of motion in order to find v

Part II :- As we know the values of u, t , g we can use the second equation of motion in order to find s.

Velocity of the ball after 5 seconds

Distance covered by the ball in 5 sec

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A group of students is investigating whether the smoothness of surface affects the force of friction. Each student pushes a book

Pani-rosa [81]

To find the average (aka mean) of a group of numbers, all we need to do is add them up then divide that number by how many numbers we started with

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90 ÷ 4 = 22.5

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In conclusion, the average distance the book traveled on ice is greater than the average distance the book traveled on concrete

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3 years ago

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Explain the relationship between air resistance and gravitational acceleration on a falling object

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Air resistance, also called drag, acts upon a falling body by slowing the body down to thr point where it stops accelerating, and it falls at a constant speed, known as the terminal volocity of a falling object. Air resistance depends on the cross sectional area of the object, which is why the effect of air resistance on a large flat surfaced object is much greater than on a small, streamlined object.

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3 years ago

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

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4 years ago

0.0084x91 ...................

Drupady [299]

Answer:

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Explanation:

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3 years ago

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Darya [45]

Answer:

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Explanation:

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3 years ago