Motion - Physics

First, we need to find the length of the pendulum. Begin with the equation for finding the period of a pendulum:

solve for to get:

Now we can plug in our given values:

Since we have the length of the pendulum determined, we can now find the period of the pendulum on Mars:

The angular frequency of a simple pendulum is , where is the length of the pendulum.

We know that T is the period. The equation for T is for harmonic motion.

Solve for by dividing the equation by on both sides. The result is , which is the answer.

We know that the equation for the period of a simple pendulum is . This equation does not depend on mass. It is only affected by the length of the pendulum (L) and the gravitational constant (g). Therefore, adding mass to the pendulum will not effect the period, so the period remains the same.

Since we are solving for string tension, we need to use the frequency equation with the tension variable in it. That equation is where is the frequency, is the number of the harmonic, is the length of the string, is the linear density of the string, and is the tension of the string.

We are given:

Next we must convert the length of to meters which is and the mass density of to . Then we plug in our known values into the equation and solve for the string tension. The result is .

For this question, we need to recall what simple harmonic motion is. Remember that it is a periodic motion where the restoring force depends on the displacement of the object undergoing these motions. So to answer this question, we need to keep this idea in mind and see which example doesn't match up.

A mass on a pendulum moving back and forth is clearly an example of simple harmonic motion. As the mass moves further from the center in either direction, it experiences a greater and greater force in the opposite direction.

A child swinging on a swing set is another correct example. This situation is analogous to the mass on a pendulum swinging back and forth.

A vibrating guitar string is yet another example of simple harmonic motion. After it is plucked, the string oscillates back and forth.

Finally, a book falling to the ground does not represent harmonic motion. Once the book is released from rest, we intuitively know that it will fall to the ground and will then stay there; in no way is there any periodic motion.

There is 90 feet between each base and distance does not depend on direction. Distance is scaler therefore, the batter travels 360 feet.

Displacement is a vector. Therefore, magnitude and direction matters and because direction matters the total displacement is 0 feet.

Displacement is a vector. Magnitude and direction matter.

We are dealing with motion in one plane and the initial velocity is zero. We must use the following kinematics equation to plug in known values and solve:

Velocity is a vector so direction matters. Therefore this creates a right triangle with the two isosceles side having a velocity of . Therefore, the hypotenuse is when you perform Pythagorean theorem.

This question requires an understanding of motion in two dimensions. The most important concept in this question is that the motion in each dimension is independent. Since the rock's initial velocity is purely in the horizontal direction, the initial velocity has no impact on the vertical velocity at any point. Likewise, the vertical acceleration has no impact on the horizontal speed. The rock will travel at in the horizontal direction throughout its entire trajectory.

To solve this problem, the first step is to set up an expression for the horizontal and vertical velocities as a function of time:

Thus,

Substituting for and solving for yields .

We first must find the initial vertical velocity () using:

We know that , since in 2 dimensions the vertical velocity at its max height is equal to zero.

we also know that because that information is given, and that

So plugging in what we know:

Knowing that is constant and equals , we can use the pythagorean theorem to determine the Resultant initial Velocity:

First let's make a table of what we know:

because it has zero velocity right when it is dropped.

It should also be noted it is simpler to define things so that the initial height is zero, so we don't have to deal with a bunch of negative numbers.

Also, since it is dropped vertically and we are only interested in the height it is dropped from we aren't interested in any information regarding the projectile's horizontal motion.

Let's use the equation:

since

We can solve for , then plug in what we know:

This is our final answer.

First let's write down the information we're given:

, since at a projectile's max height its vertical velocity equals zero.

Let's use the equation

to solve for the time it takes the projectile to reach its max height.

Solving for :

Now lets plug in what we know and calculate the time:

, which is our final answer.

First let's write down the information we're given:

, since at a projectile's max height its vertical velocity equals zero.

Let's use the following equation to solve for the time it takes the projectile to reach its max height:

Solving for :

Now lets plug in what we know and calculate the time:

Since the projectile takes twice as long to land as its does to reach its max height we simply multiple the we found by , which gives us .

First let's write out the information we are given:

We know that in order to obtain the time it takes a projectile to hit the ground we just multiply by , since the time it takes to reach the max height is half of the total time the project is in the air, so

Now we can use the equation

Since we are only concerned with the particle's motion in the horizontal direction, and we know that the horizontal velocity is constant . We also know since that is the definition of the initial position.

This gives us:

Let's plug in what we have:

. This is the range and our final answer. The reason why we can just use as the horizontal velocity is because the project is launched at a ° angle.

This is a projectile motion problem. The problem states that it was thrown off the building at an angle of 30 degrees to the horizontal. So before we can find time, we need to find the horizontal and vertical parts of this velocity.

Since and

It follows that

Now, before we can find far it went, we need to find how long it was in the air. We need to solve for time.

With the equation we can find t.

X is the initial height which in this case is the tall building. V is the initial vertical velocity and

Plugging all that in and solving yields

Knowing that the time the ball is in the air is and the constant horizontal velocity is , we can plug in these known values into the simple distance formula to solve for distance.

There are many explanations for this. First you could simply insert a velocity at all of these angles and see which ends up with the greatest change in horizontal distance. You also could use basic calculus and solve for the greatest theta.