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.

$\displaystyle v_f^2=v_i^2+2a\Delta y$

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

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

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

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

$\displaystyle 797.2m=\Delta y$

To find the time for the projectile to reach its maximum height, we can use the appropriate kinematics equation:

$\displaystyle v_f=v_i+at$

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 time.

$\displaystyle v_{fy}=v_{iy}+at$

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

$\displaystyle -125\frac{m}{s}=-9.8\frac{m}{s^2}t$

$\displaystyle \frac{-125\frac{m}{s}}{-9.8\frac{m}{s^2}}=t$

$\displaystyle 12.76s=t$

To find the total time, we can solve for the time to reach the peak and then double this value to find the total time.

To find the time for the projectile to reach its maximum height, we can use the appropriate kinematics equation:

$\displaystyle v_f=v_i+at$

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 time.

$\displaystyle v_{fy}=v_{iy}+at$

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

$\displaystyle -125\frac{m}{s}=-9.8\frac{m}{s^2}t$

$\displaystyle \frac{-125\frac{m}{s}}{-9.8\frac{m}{s^2}}=t$

$\displaystyle 12.76s=t$

Double this value to find the total flight time.

$\displaystyle 2(12.76s)=25.5s$

To solve for the horizontal distance we can use the simple formula for velocity, since there is no acceleration in the horizontal direction.

$\displaystyle \Delta x =v_x*t$

This leaves us two distinct variables: $\displaystyle v_x$ and $\displaystyle t$.

To find our $\displaystyle v_x$, we need to use cosine.

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

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

$\displaystyle 216.5\frac{m}s=v_x$

To find the time, we can solve for the time to reach the peak and then double this value to find the total time.

To find the time for the projectile to reach its maximum height, we can use the appropriate kinematics equation:

$\displaystyle v_f=v_i+at$

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 time.

$\displaystyle v_{fy}=v_{iy}+at$

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

$\displaystyle -125\frac{m}{s}=-9.8\frac{m}{s^2}t$

$\displaystyle \frac{-125\frac{m}{s}}{-9.8\frac{m}{s^2}}=t$

$\displaystyle 12.76s=t$

Double this value to find the total flight time.

$\displaystyle 2(12.76s)=25.5s$

Now that we know our total time and horizontal velocity, we can return to the velocity equation to solve for the distance traveled.

$\displaystyle \Delta x =v_x*t$

$\displaystyle \Delta x =216.5\frac{m}{s}*25.52s$

$\displaystyle \Delta x =5525.08m$

Start by drawing a picture.

By traveling both north and east, Peter has displacement along both the x-axis and the y-axis. Recognize that this makes a right triangle.

We can solve for the y-axis displacement by using trigonometry.

$\displaystyle \sin(\Theta)=\frac{\text{opposite}}{\text{hypotenuse}}$

In this case, we know the angle and the hypotenuse. The y-displacement is opposite the angle.

$\displaystyle \sin(38.3^o)=\frac{Y}{30.7m}$

Multiply both sides by $\displaystyle 30.7$ and solve.

$\displaystyle (30.7m)\sin(38.3^o)=Y$

$\displaystyle 19.03m=Y$

Since we are solving for the displacement, we need to include the direction. The displacement is $\displaystyle \small 19.03m\ \text{north}$.