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vazorg [7]
4 years ago
7

A cannonball is fired horizontally at the same time a ball being dropped from the same height How do the times it takes them to

hit the ground compare?
Physics
1 answer:
ipn [44]4 years ago
3 0

Answer:

The cannonball and the ball will both take the same amount of time before they hit the ground.

Explanation:

For a ball fired horizontally from a given height, there is only a vertical acceleration on it towards the ground. This acceleration is equal to the acceleration due to gravity (g = 9.81 m/s^2). A ball dropped from a height will also only experience the same vertical acceleration downwards which is also equal to g = 9.81 m/s^2.

Therefore both the cannonball and the ball will take the same amount of time to hit the ground if they are released/fired from the same height.

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3 years ago
A particle moves along a straight line with an acceleration of a = 5>(3s 1>3 + s 5>2) m>s2, where s is in meters. De
Len [333]

Answer:

v=1.295

Explanation:

What we are given:

a=5÷(3s^(1/3)+s^(5/2)) m/s^2

Start by using equation  a ds = v dv

This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:

\int\limits^a_b {x} \, dx

<em>a=2</em>

<em>b=1</em>

<em>x=5÷(3s^(1/3)+s^(5/2)) </em><em>m/s^2</em>

<em>dx=dv</em>

Integrate the left side the standard method.

\int\limits^a_b {x} \, dx

<em>a=v</em>

<em>b=0</em>

<em>dx=dv</em>

<em>Integrating</em>

=v^2/2

Use Simpson's rule for the right site.

\int\limits^a_b {x} \, dx

<em>a=b</em>

<em>b=a</em>

<em>x=f(x)</em>

f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)

If properly applied. you should now have the following equation:

v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]

        =0.8376

Solve for v.  

      v=1.295

5 0
3 years ago
True or False? The dependent variable value changes according to changes in the other variables.
Maru [420]
True because when the independent answer changes, then the dependent variable, which relies on the independent variable, changes.
3 0
3 years ago
Read 2 more answers
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