In order to find the distance that the ball travels horizontally one must multiply the velocity of the ball by the amount of time that the ball is traveling. Time in the air can be found by figuring out the amount of time that it takes for the ball to hit the ground relative to the vertical distance. Using the equation:

 $\displaystyle d=V_{0}t+\frac{1}{2}at^{^{2}}$ 

where d is $\displaystyle 37m$ (height of the cliff plus height of the man), $\displaystyle V_{0}$ is equal to zero because there is no initial vertical velocity, and acceleration is equal to the gravity constant, time is found to be $\displaystyle 2.75s$. In order to find distance traveled horizontally, one can use the equation:

$\displaystyle d=V_0t+\frac{1}{2}at^2$

Because the vertical and horizontal axis are completely independent of each other there is no acceleration in the horizontal direction, the equation then becomes simplified to:

$\displaystyle d=V_{0}t$

After plugging in numbers

$\displaystyle d=36*2.75$

$\displaystyle d=99m$

If $\displaystyle h=0$, the horizontally thrown ball will travel no distance. As the height of the cliff increases, the horizontally thrown ball will have more and more air time until it begins to travel further than the ball thrown $\displaystyle 45^{\circ}$ above the horizontal. This is because the distance each ball will go equals the time it's in the air multiplied by its horizontal velocity (since it's not accelerating in the x-direction). The time the ball thrown $\displaystyle 45^{\circ}$ above the horizontal is always the same amount greater than the time the ball thrown horizontally is in the air. The ball thrown horizontally has a horizontal velocity greater than the ball thrown $\displaystyle 45^{\circ}$ above the horizontal (by a factor of $\displaystyle \sqrt{2}$). This means the ball thrown horizontally travels a distance of $\displaystyle v*t$ while the ball thrown $\displaystyle 45^{\circ}$ above the horizontal travels a distance of $\displaystyle \frac{\sqrt2v(t+k)}{2}$. If we equate these two expressions, we get $\displaystyle t=(1+\sqrt{2})k$. When $\displaystyle t$ is less than this, the ball thrown $\displaystyle 45^{\circ}$ above the horizontal travels further and when $\displaystyle t$ is greater the ball thrown horizontally travels further. $\displaystyle t=(1+\sqrt{2})k$ represents the time is takes for the ball thrown horizontally to hit the ground when the height of this cliff is $\displaystyle h_{equal}$.

Because the mass of object 1 is given, the force acting on it must be found in order to figure out how fast it is accelerating. These 3 variables are related by the equation:

$\displaystyle F=ma$

Because there is no friction in the system, the only force acting on the object is the tension in the rope. The magnitude of the force of tension is the magnitude of the gravitational force of mass 2. Gravitational force is equal to:

$\displaystyle F=mg$

where g is the gravity constant which is equal to $\displaystyle 9.81\frac{m}{s^{2}}$ which is also the acceleration of this object. Since this force is the only force acting on the object, accelerating can be found relating the 2 equations above as follows:

$\displaystyle m_{2}g=m_{1}a_{1}$

After plugging in values, the acceleration of mass 1 is found to be:

$\displaystyle 6.5*9.81=4.0a_{1}$

$\displaystyle a_{1}\approx 16\ \frac{m}{s^{2}}$

To solve this question, we'll first need to break up the velocity into its component parts. For this question, we're only concerned about motion in the vertical direction, so we'll first need to find the initial velocity in the vertical direction.

$\displaystyle v_{yi}=v_{i}sin(\Theta )$

$\displaystyle v_{yi}=(50\: \frac{m}{s})sin(30^{o})=25\: \frac{m}{s}$

Now that we know what the initial velocity is, we can determine what the final velocity will be after $\displaystyle 2\: s$.

$\displaystyle \frac{\Delta v}{t}=a$

$\displaystyle \frac{v_{yf}-v_{yi}}{t}=a$

$\displaystyle v_{yf}=v_{yi}+at$

$\displaystyle v_{yf}=25\: \frac{m}{s}+(-9.8\: \frac{m}{s^{2}})(2\:sec)$

$\displaystyle v_{yf}=25\: \frac{m}{s}-19.6\: \frac{m}{s}=5.4\: \frac{m}{s}$

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

$\displaystyle \Delta y = v_{i}+\frac{1}{2}at^2$

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

$\displaystyle t = \sqrt{\frac{2\Delta y}{a}}$

Plugging in our values, we get:

$\displaystyle t = \sqrt{\frac{2(1.2m)}{10\frac{m}{s^2}}}$

$\displaystyle t = 0.49s$

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

$\displaystyle d = \left ( 17\frac{m}{s}\right )(0.49s) = 8.3m$

First, let's separate the given velocity into its vertical and horizontal components:

$\displaystyle v_x = vcos(\theta)$

$\displaystyle v_y = vsin(\theta)$

$\displaystyle v_x= \left ( 30\frac{m}{s}\right )cos(15^{\circ})$

$\displaystyle v_y= \left ( 30\frac{m}{s}\right )sin(15^{\circ})$

$\displaystyle v_ x= 28.97777\frac{m}{s}$

$\displaystyle v_y = 7.76457\frac{m}{s}$

Since we are neglecting air resistance, we can easily calculate how long it takes for the ball to reach the second player:

$\displaystyle t = \frac{d}{v_x}=\frac{25m}{28.97777\frac{m}{s}} = 0.86273s$

Now we can use this time to calculate the height of the ball when it reaches the second player:

$\displaystyle y_f = y_i+v_{y_i}t+\frac{1}{2}at^2$

Plugging in our values, we get:

$\displaystyle y_f = 1m+\left ( 7.76457\frac{m}{s}\right )(0.86273s)+\frac{1}{2}\left ( -10\frac{m}{s^2}\right )(0.86273s)^2$

$\displaystyle y_f=3.98m$

This problem is greatly simplified since the home run fence has the same height as the initial height of the ball. Therefore, we can use this range equation to solve this problem in one calculation: 

$\displaystyle R = \frac{v_o^2sin(2\theta)}{g}$

Rearranging for velocity, we get:

$\displaystyle v_o = \sqrt{\frac{R\cdot g}{sin(2\theta)}}$

Plugging in our values, we get:

$\displaystyle v_o = \sqrt{\frac{(120m)\left ( 10\frac{m}{s^2}\right )}{sin(100^{\circ})}}$

$\displaystyle v_o = 34.9\frac{m}{s}$

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:

 $\displaystyle d = v_x\cdot t$

Rearranging for time:

$\displaystyle t = \frac{d}{v_x}$

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

$\displaystyle v_x = vcos(\theta)$

$\displaystyle t = \frac{d}{vcos(\theta)}$

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

$\displaystyle \Delta y= v_{y_i}t+\frac{1}{2}at^2$

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

$\displaystyle \Delta y= vsin(\theta)t+\frac{1}{2}at^2$

Plugging in our expression for time, we get:

$\displaystyle \Delta y=vsin(\theta)\left (\frac{d}{vcos(\theta)} \right )+\frac{1}{2}a\left ( \frac{d}{vcos(\theta)}\right )^2$

Now let's simplify this a bit:

$\displaystyle \Delta y =d\cdot tan(\theta)+\frac{a\cdot d^2}{2v^2cos^2(\theta)}$

Rearranging:

$\displaystyle \Delta y -d\cdot tan(\theta)=\frac{a\cdot d^2}{2v^2cos^2(\theta)}$

And some more:

$\displaystyle 2v^2cos^2(\theta)=\frac{a\cdot d^2}{\Delta y -d\cdot tan(\theta)}$

One last time:

$\displaystyle v = \sqrt{\frac{a\cdot d^2}{\left ( \Delta y - d\cdot tan\theta\right )\left ( 2cos^2\theta\right )}}$

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

$\displaystyle v = \sqrt{\frac{\left ( 10\frac{m}{s^2}\right )(120m)^2}{\left ( 4m-(120m)tan(50^\circ)\right )(2cos^2(50^{\circ}))}}$

$\displaystyle v_i = 35.4\frac{m}{s}$

From the problem statement, we can use the following piecewise function to express the effective "height" at any point within the stadium:

If   $\displaystyle x < 100m$   then   $\displaystyle y = 0m$

If   $\displaystyle x\geq 100m$   then   $\displaystyle y = 5m+(x-100)sin(30^{\circ})$

Before, moving on, make sure you understand the reasoning of this function. The wall begins at a distance of $\displaystyle 100m$, a height of $\displaystyle 5m$, and rises at an angle of $\displaystyle 30^{\circ}$.

Using the total horizontal distance traveled, we can calculate the ball's final height using the above expression:

$\displaystyle y = 5m+(120m-100m)sin(30^{\circ}) = 15m$

Now we can use the following two kinematics expression:

$\displaystyle \Delta y = v_{y_{i}}t+\frac{1}{2}at^2$

$\displaystyle t = \frac{d}{v_{x_{i}}}$

Where d is the horizontal distance traveled. Replacing out component velocities, we get:

$\displaystyle \Delta y = v_isin(\theta)t+\frac{1}{2}at^2$

$\displaystyle t = \frac{d}{v_icos(\theta)}$

Substituting our expression for time into the other expression, we get:

$\displaystyle \Delta y = dtan(\theta)+\frac{1}{2}a\left ( \frac{d}{v_icos(\theta)}\right )^2$

Simplifying, we get:

$\displaystyle \Delta y = dtan(\theta)+\frac{ad^2}{2v_i^2cos^2(\theta)}$

Rearranging:

$\displaystyle \frac{ad^2}{2v_i^2cos^2(\theta)} = dtan(\theta)-\Delta y$

And again:

$\displaystyle \frac{ad^2}{dtan(\theta)-\Delta y} = 2v_i^2cos^2(\theta)$

One last time:

$\displaystyle v_i = \sqrt{\frac{ad^2}{2cos^2(\theta)\left (dtan(\theta)-\Delta y \right )}}$

It's nice and messy, but we have all of these values, so time to plug and chug:

$\displaystyle v_i = \sqrt{\frac{\left ( 10\frac{m}{s^2}\right )(120m)^2}{2cos^2(50^{\circ})((120m)tan(50^{\circ})-14m)}}$

$\displaystyle v_i = 36.8\frac{m}{s}$

From the problem statement, we can use the following piecewise function to express the effective "height" at any point within the stadium:

If   $\displaystyle x < 100m$   then   $\displaystyle y = 0m$

If   $\displaystyle x\geq 100m$   then   $\displaystyle y = 5m+(x-100)sin(30^{\circ})$

Before, moving on, make sure you understand the reasoning of this function. The wall begins at a distance of $\displaystyle 100m$, a height of $\displaystyle 5m$, and rises at an angle of $\displaystyle 30^{\circ}$.

We can use the following expression to calculate the final height of the ball:

$\displaystyle y_f= y_i + v_{y_{i}}t+\frac{1}{2}at^2$

Plugging in our values, we get:

$\displaystyle y_f = 1m + \left ( 25\frac{m}{s}\right )(4s)+\frac{1}{2}\left ( -10\frac{m}{s^2}\right )(4s)^2$

$\displaystyle y_f = 21m$

Then we can use the piecewise expression to determine to total horizontal distance traveled:

$\displaystyle 21m = 5m+ (x-100m)sin(30^{\circ})$

$\displaystyle x = \frac{16m}{sin(30^{\circ})}+100m$

$\displaystyle x = 132m$