The man's fall will be parabolic as there will be both horizontal and vertical components. His vertical component of the fall will be standard free-fall caused by his acceleration due to gravity. His horizontal component of the fall will come from him "leaning too far" in one direction. Even a small horizontal velocity will create a horizontal trajectory.

This is why when people lean and fall off of ladders they either try to grab onto the ladder (try to negate their horizontal velocity) or fall a small distance away from the base of the ladder.

Even though the problem gives us an initial velocity, we need to break it down into horizontal and vertical components.

We can plug in the given values and find the vertical velocity.

Remember that the vertical velocity at the highest point of a parabola is zero. Now that we know the initial and final vertical velocities, we can plug our values into an equation to solve for the maximum height.

Remember, only tells us the CHANGE in the vertical direction. The rock started at the top of a tall building, then rose an extra .

Its highest point is above the ground.

The key to solving this problem is to recognize that in two dimensional motion, the x-direction and the y-direction are dealt with separately. The question asks how long the projectile is in the air so we are dealing only with motion in the y-axis.

Since the projectile was fired at an angle we must use trigonometry to find the velocity in the y-axis . Then we use a kinematic equation to solve for time.

Initial velocity in the y-axis:

Now we will find the time it takes for the projectile to hit the ground using a kinematic equation:

We use at the top of the projectile's trajectory, where its velocity is changing from going up, to going down.

Note that this is the time it takes for the projectile to reach maximum height. The time it takes to get back to the ground from here is the same as the time it took to get here. Thus, to find the total flight time, we must double our answer.

The given velocity won't help us much here. We need to break it down into horizontal and vertical components.

Use the sine function with the initial velocity and angle to find the vertical velocity.

Now we need to work on finding the time in the air. To do this, we need to break the rock's path into two parts. The first part is the time that the rock is rising to its maximum height, and the second part is the time that it is falling from the highest point to the ground.

For the first part, we can assume that the final vertical velocity is zero, since this will be the top of the parabola. Using the initial velocity, final velocity, and gravity, we can solve for the time to travel this portion of the path.

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This is the time that the rock is traveling upward. Now we need to focus on the time that the rock travels downward. We don't know the final velocity at the end of the parabola, so we can't use the equation from the first part. If we can find the total height of the parabola, we can use a different equation to solve for time.

Remember that the vertical velocity at the highest point of a parabola is zero. Now that we know the initial vertical velocity, we can plug this into an equation to solve for the distance that the rock travels from its maximum height to the original height.

Remember, only tells us the CHANGE in height. Since the rock started at the top of a tall building, if it rose an extra , then at its highest point it is above the ground.

This means that our for the second equation will be (the change in height is negative because it travels downward.) Use this total distance and the velocity at the top of the peak (zero) to solve for the time that the rock travels down.

Finally, add the two times together to find the total time in flight.

The given initial velocity is at an angle, so we have to use some trigonometric functions to break it into horizontal and vertical components.

Effectively, the initial velocity becomes the hypotenuse of a right triangle with the horizontal velocity becoming the base and the vertical velocity becoming the height. To find the vertical velocity, we use the relationship between the hypotenuse and the opposite side.

Use the given initial velocity and angle to solve for the vertical velocity.

The problem gives us the initial horizontal velocity. This velocity will only affect the distance the ball travels in the horizontal direction; it will have no effect on the time the ball is in the air. Since there is no angle of trajectory, the ball has no initial vertical velocity.

We know the height of the table, the initial velocity, and gravity. Using these values with the appropriate motion equation, we can solve for the time.

The best equation to use is:

We can use our values to solve for the time. Keep in mind that the displacement will be negative because the ball is traveling in the downward direction!

The given initial velocity is at an angle, so we have to use some trigonometric functions to break it into horizontal and vertical components.

Effectively, the initial velocity becomes the hypotenuse of a right triangle with the horizontal velocity becoming the base and the vertical velocity becoming the height. To find the horizontal velocity, we use the relationship between the hypotenuse and the adjacent side.

Use the given initial velocity and angle to find the horizontal velocity.

We can solve for the horizontal distance using only the horizontal velocity: .

We are given the value of , but we need to find the time. Time in the air will be determined by the vertical components of the ball's motion.

We know the height of the table, the initial velocity, and gravity. Using these values with the appropriate motion equation, we can solve for the time.

The best equation to use is:

We can use our values to solve for the time. Keep in mind that the displacement will be negative because the ball is traveling in the downward direction!

Now we have both the time and the horizontal velocity. Use the original equation to solve for the distance.

This problem involves understanding motion in two dimensions: horizontal (we will use ) and vertical (we will use ).

Let's begin by writing down what we know, and what we need to find.

What we know:

What we need to find:

To find velocity, we need to know both the horizontal and the vertical components of the ball's velocity. We will break them down below.

The horizontal component of the velocity, , does not change with time so we can solve for that simply by finding the horizontal distance, , covered over time .

The vertical component of the velocity, , will be slightly more challenging to find. Because gravity acts in the vertical direction, we now need to take acceleration into account. Because of this, we will try to find the initial vertical velocity, , of the ball. We are told in the problem that the ball hits the scoreboard at the top of its trajectory. At the top of the trajectory, vertical velocity will be zero. This also marks the midpoint of the ball's flight, or half the given time. With this information, we can solve for the initial vertical velocity:

Now that we have both the horizontal and vertical components, we can use the Pythagorean theorem to solve for the total initial velocity, .

The question asks how far the projectile can travel before landing. Since we deal with the x and y-axes of 2-dimensional motion separately, it is important to recognize that only the movement in the x-axis is relevant. If you are driving 50mph and do so for three hours you have driven 150mi. Similarly, our projectile is traveling at a certain speed horizontally and does so for 20s. Thus we simply multiply its horizontal speed by the time the projectile was in the air. Since the projectile was launched at an angle we use trigonometry to find its horizontal speed:

Now use the formula for range and plug in the velocity in the x-direction (constant) to find the distance travelled.