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:

\(\displaystyle \Delta y =v_it+\frac{1}{2}at^2\)

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!

\(\displaystyle -3m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -3m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -3m=(-4.9\frac{m}{s^2})t^2\)

\(\displaystyle \frac{-3m}{-4.9\frac{m}{s^2}}=t^2\)

\(\displaystyle 0.612s^2=t^2\)

\(\displaystyle \sqrt{0.612s^2}=\sqrt{t^2}\)

\(\displaystyle 0.782s=t\)

One of the key concepts of parabolic motion in freefall is that the horizontal velocity, \(\displaystyle v_x\), does not change. In order for velocity to change, there must be an acceleration. If an object is accelerating, then there must be a force acting on the object. The only force on the object during freefall is gravity, which acts in the vertical direction and cannot affect the horizontal velocity.

Since our given \(\displaystyle v_{ix}=19\frac{m}{s}\), our \(\displaystyle v_{fx}=19\frac{m}{s}\).

In order for us to find the final angle, we need to find the final horizontal and vertical components.

The only force acting on the ball while in the air is gravity. Since gravity acts in the vertical direction, there is no force in the horizontal direction. This means that the horizontal velocity will not change.

\(\displaystyle v_{xi}=v_{xf}=19\frac{m}{s}\)

The problem states that the initial velocity is only in the horizontal direction; the initial vertical velocity is zero. We now know initial velocity, acceleration, and distance traveled.

Remember, even though the distance it will travel is \(\displaystyle 3m\), its displacement will be \(\displaystyle -3m\) as it moves in the downward direction.

Using these values and the appropriate motion equation, we can solve for the final velocity. The best equation to use is:

\(\displaystyle v_{fy}^2=v_{iy}^2+2a\Delta y\)

Use the given values to find the final velocity.

\(\displaystyle v_{fy}^2=0\frac{m}{s}^2 +2(-9.8\frac{m}{s^2})(-3m)\)

\(\displaystyle v_{fy}^2=58.8\frac{m^2}{s^2}\)

\(\displaystyle \sqrt{v_{fy}^2}}=\sqrt{58.8\frac{m^2}{s^2}}\)

\(\displaystyle v_{fy}=7.67\frac{m}{s}\)

Now that we know the horizontal and vertical components of the final velocity, we can solve for the angle by considering the components are legs of a right triangle. Using trigonometric identities, we can solve for the angle.

 \(\displaystyle \tan(\Theta)=\frac{v_y}{v_x}\)

Using our horizontal and vertical velocities, we can calculate the angle.

\(\displaystyle \tan(\Theta)=\frac{7.67\frac{m}{s}}{19\frac{m}{s}}\)

\(\displaystyle \tan(\Theta)=0.404\)

\(\displaystyle \Theta=\tan^{-1}(0.404)\)

\(\displaystyle \Theta=22^\circ\)

In order for us to find the final resultant speed, we need to find the final horizontal and vertical components.

The only force acting on the ball while in the air is gravity. Since gravity acts in the vertical direction, there is no force in the horizontal direction. This means that the horizontal velocity will not change.

\(\displaystyle v_{xi}=v_{xf}=19\frac{m}{s}\)

The problem states that the initial velocity is only in the horizontal direction; the initial vertical velocity is zero. We now know initial velocity, acceleration, and distance traveled.

Remember, even though the distance it will travel is \(\displaystyle 3m\), its displacement will be \(\displaystyle -3m\) as it moves in the downward direction.

Using these values and the appropriate motion equation, we can solve for the final velocity. The best equation to use is:

\(\displaystyle v_{fy}^2=v_{iy}^2+2a\Delta y\)

Use the given values to find the final velocity.

\(\displaystyle v_{fy}^2=0\frac{m}{s}^2 +2(-9.8\frac{m}{s^2})(-3m)\)

\(\displaystyle v_{fy}^2=58.8\frac{m^2}{s^2}\)

\(\displaystyle \sqrt{v_{fy}^2}}=\sqrt{58.8\frac{m^2}{s^2}}\)

\(\displaystyle v_{fy}=7.67\frac{m}{s}\)

Now that we know the final velocity in both the horizontal and vertical directions, we can use the Pythagorean theorem to solve for the total velocity.

\(\displaystyle v_{resultant}^2=v_x^2+v_y^2\)

\(\displaystyle v_{resultant}^2=(19\frac{m}{s})^2+(7.67\frac{m}{s})^2\)

\(\displaystyle v_{resultant}^2=361\frac{m}{s}+58.83\frac{m}{s}\)

\(\displaystyle v_{resultant}^2=419.83\frac{m}{s}\)

\(\displaystyle \sqrt{v_{resultant}^2}=\sqrt{419.83\frac{m}{s}}\)

\(\displaystyle v_{resultant}=20.49\frac{m}{s}\)

The problem states that the initial velocity is only in the horizontal direction; the initial vertical velocity is zero. We now know initial velocity, acceleration, and distance traveled.

Remember, even though the distance it will travel is \(\displaystyle 3m\), its displacement will be \(\displaystyle -3m\) as it moves in the downward direction.

Using these values and the appropriate motion equation, we can solve for the final velocity. The best equation to use is:

\(\displaystyle v_{fy}^2=v_{iy}^2+2a\Delta y\)

Use the given values to find the final velocity.

\(\displaystyle v_{fy}^2=0\frac{m}{s}^2 +2(-9.8\frac{m}{s^2})(-3m)\)

\(\displaystyle v_{fy}^2=58.8\frac{m^2}{s^2}\)

\(\displaystyle \sqrt{v_{fy}^2}}=\sqrt{58.8\frac{m^2}{s^2}}\)

\(\displaystyle v_{fy}=7.67\frac{m}{s}\)

Because we just took the square root of a number, we got an absolute value for our \(\displaystyle v_{fy}\); however, velocity is a vector and can be either positive or negative depending on direction. Because the ball is headed downward, the final velocity should correctly be \(\displaystyle -7.67\frac{m}{s}\). Remember that a negative number squared gives a positive value, just like a positive number.

We can solve for the horizontal distance using only the horizontal velocity: \(\displaystyle \Delta x =v_xt\).

We are given the value of \(\displaystyle v_x\), 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:

\(\displaystyle \Delta y =v_it+\frac{1}{2}at^2\)

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!

\(\displaystyle -3m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -3m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -3m=(-4.9\frac{m}{s^2})t^2\)

\(\displaystyle \frac{-3m}{-4.9\frac{m}{s^2}}=t^2\)

\(\displaystyle 0.612s^2=t^2\)

\(\displaystyle \sqrt{0.612s^2}=\sqrt{t^2}\)

\(\displaystyle 0.782s=t\)

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

\(\displaystyle \Delta x =v_xt\)

\(\displaystyle \Delta x =(19\frac{m}{s})(0.782s)\)

\(\displaystyle \Delta x =14.86m\)

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.

Remember that the velocity in the horizontal direction stays constant through the projectile's motion. There is no force in the horizontal direction, only in the vertical direction. That means the initial and final horizontal velocities will be the same.

To find our \(\displaystyle v_x\), we need to use cosine trigonometry, with the horizontal velocity as the adjacent side and the total initial velocity as the hypotenuse.

\(\displaystyle \cos(\theta)=\frac{adj}{hyp}=\frac{v_x}{v_i}\)

\(\displaystyle v_i*\cos(\theta)=v_x\)

\(\displaystyle 250\frac{m}{s}*\cos(30^\circ)=v_x\)

\(\displaystyle 216.5\frac{m}s=v_x\)

We are given the total initial velocity and the angle of the initial trajectory. Using these values, we can use trigonometry to solve for the initial vertical velocity.

We will need to use sine, with the vertical velocity as the opposite side and the total velocity as the hypotenuse.

\(\displaystyle \sin(\theta)=\frac{opp}{hyp}=\frac{v_y}{v_i}\)

\(\displaystyle v_i*\sin(\theta)=v_y\)

\(\displaystyle 250\frac{m}{s}*\sin(30^\circ)=v_y\)

\(\displaystyle 125\frac{m}s=v_y\)

To find the height of the projectile, we can use the appropriate kinematics equation:

\(\displaystyle v_f^2=v_i^2+2a\Delta y\)

We know that the final velocity at the maximum height will be zero, and we also know the acceleration due to gravity. Before we can use the equation, however, we must solve for the initial vertical velocity. We are given the total initial velocity and the angle of the initial trajectory. Using these values, we can use trigonometry to solve for the initial vertical velocity.

\(\displaystyle v_i*\sin(\theta)=v_y\)

\(\displaystyle 250\frac{m}{s}*\sin(30^\circ)=v_y\)

\(\displaystyle 125\frac{m}s=v_y\)

Now that we know the initial vertical velocity, we can return to the kinematics equation to solve for the final displacement.

\(\displaystyle v_f^2=v_i^2+2a\Delta y\)

\(\displaystyle (0\frac{m}{s})^2=(125\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y\)

\(\displaystyle 0=15625\frac{m^2}{s^2}+(-19.6\frac{m}{s^2})\Delta y\)

\(\displaystyle -15625\frac{m^2}{s^2}=-19.6\frac{m}{s^2}*\Delta y\)

\(\displaystyle \frac{-15625\frac{m^2}{s^2}}{-19.6\frac{m}{s^2}}=\Delta y\)

\(\displaystyle 797.2m=\Delta y\)