The correct answer is . This vector indicates that in order to get to point B from point A you must be 2 units to the right - the first number in vector notation indication horizontal movement or movement in the x-direction - and 5 units upward - the second number in vector notation indicating vertical movement or movement in the y-direction.

The velocity must be broken down into x (horizontal) and y (vertical) components. We can use the y component to find how high the object gets. To find vertical velocity, vy, use .

Next we find how long it takes to reach the top of its trajectory using .

t = 5.3s

Finally, find how high the object goes with .

There are multiple ways to attack this problem, so don't worry if you took a different route. However, no matter which way you go, you'll be substituting small expressions into larger ones.

First, let's develop an expression of how long it takes the ball to reach the home run fence. We know that:

Rearranging for time:

Then we can substitute the following expression in for horizontal velocity:

Now that we have an expression for time, we can use the following expression:

Plugging in an expression for initial vertical velocity, we get:

Plugging in our expression for time, we get:

Now let's simplify this a bit:

Rearranging:

And some more:

One last time:

It may look a bit nasty, but we know all of these values, so time to plug and chug:

First, we can calculate how long it takes for the ball to hit the ground using the following expression:

Since the original vertical velocity is 0, we can rearrange to get:

Plugging in our values, we get:

Multiplying this by the velocity given in the problem statement (horizontal velocity), we get:

First, find the horizontal (x) and vertical (y) components of the velocity

Next, find how long the object is in the air by calculating the time it takes it to reach the top of its path, and doubling that number.

t = 6.25s

Total time in the air is therefore 12.5s (twice this value).

Finally, find distance traveled my multiplying horizontal velocity and time.

We can start this problem by determining how much time it takes the box the reach the ground. Since the vertical distance is 30m and we know the angle of the ramp.

Now that we have the distance traveled, we can determine the acceleration on the box due to gravity, using the equation .

We can plug these values into the following distance equation and solve for time.

Now that we know the acceleration on the box and the time of travel, we can use the equation to solve for the velocity.

The cannon ball has a two-dimensional motion, specifically projectile motion. Any two-dimensional motion consists of two one-dimensional motions happening at the same time. This is why we break vectors into components, so we can analyze each one-dimensional motion separately. The motion in the y-axis and the x-axis are connected by the time variable, since they happen at the same time. For projectile motion, we have that on the y-axis (the vertical axis) a constant acceleration (gravity) is acting on the cannon ball. For the y-axis motion with constant acceleration or free fall (which is motion with constant acceleration where the acceleration is gravity) are correct choices. On the x-axis (the horizontal axis), there is no acceleration acting on the cannon ball horizontally. Therefore, on the x-axis the cannon ball experiences motion with constant velocity. Then the only answer choice that describes the motion of the cannon ball in each axis correctly is: y-axis: motion with constant acceleration and x-axis: motion with constant velocity.

Determining the time for the upward velocity to equal zero:

The ball will take the same amount of time to descend back to it's initial height, the height of Jennifer.

Determining how far horizontally the ball will travel in that time.

In this question, we're told that a projectile (the clown) is launched at a certain velocity and a certain angle. We're asked to find the maximum height of the projectile.

We need to recognize that this is a question concerning motion in two dimensions, both x and y. This means that the projectile will be launched such that it travels in a parabolic path. However, for the purposes of this question, we can neglect motion in the x direction and instead direct our attention to the projectile's vertical motion.

Since the projectile travels in a projectile path, we know that at the instant it reaches its maximum height, its velocity will fall to zero. We can take advantage of this fact, combined with one of the kinematic equations (since we know acceleration due to gravity is constant), in order to find our answer.

And since we know that gravity is oriented in the downward direction, we can give it a negative value.

Now, we need to find an expression that gives us the clown's initial velocity in the vertical direction. Remember, the question stem gives us the clown's initial velocity, but this is not the same as its velocity in the y direction. The value that is given in the question stem is a combination of both the x direction velocity as well as the y direction velocity. Thus, we'll need to use the information about the clown's initial velocity provided in the question stem in order to calculate the initial vertical velocity.

Now that we have an expression for the initial vertical velocity, we can go ahead and plug this value into the prior expression that we derived.

Finally, if we plug in the known values, we can calculate our answer.

Determining horizontal component of velocity:

Using

Solving for

Combining equations

Converting to

Converting yards to meters

Plugging in values:

Using