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:

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

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

$\displaystyle x = 132m$

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 quickly determine the horizontal distance traveled by the ball:

$\displaystyle x = v_xt = \left ( 30\frac{m}{s}\right )(3.5s) = 105m$

Then we can use the piecewise function to calculate the final height of the ball. Since $\displaystyle x\geq 100m$, we will use the second expression:

$\displaystyle y = 5m + (105m-100m)sin(30^{\circ}) = 7.5m$

Then we can use the following kinematics equation:

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

Rearranging, we get:

$\displaystyle v_{y_{i}}= \frac{\Delta y}{t}-\frac{at}{2}$

Plugging in our values, we get:

$\displaystyle v_{y_{i}}=\frac{7.5m-1m}{3.5s} - \frac{\left ( -10\frac{m}{s^2}\right )(3.5s)}{2}$

$\displaystyle v_{y_{i}}=19.4357\frac{m}{s}$

Then we can use the following expression to determine the initial angle of the ball:

$\displaystyle tan(\theta_i)=\frac{v_{y_{i}}}{v_{x_{i}}}$

Rearranging and plugging in our values, we get:

$\displaystyle \theta_i = tan^{-1}\left ( \frac{19.4357\frac{m}{s}}{30\frac{m}{s}}\right )$

$\displaystyle \theta_i = 32.8^{\circ}$

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 calculate the total horizontal distance traveled by the ball using the following expression:

$\displaystyle x = v_xt = \left ( 25\frac{m}{s}\right )(3.5s)$

$\displaystyle x = 87.5$

The ball has yet to reach the home run wall, so it will land back on the ground with a final height of $\displaystyle 0m$.

We know that the brick is dropped from rest $\displaystyle (v_{0}=0)$ at the top of the building and, ignoring air friction, we know the only relevant force is due to gravity $\displaystyle (a_{g} = -9.81\frac{m}{s})$.  Using the kinematic equation $\displaystyle \Delta x=v_{0}t+\frac{1}{2}at^{2}$, plug in our known values for $\displaystyle v_{0}$, $\displaystyle a$, and $\displaystyle t$, then solve to get $\displaystyle \Delta x = 490m$.

The problem can be split into two parts: when the brick is in the air above the building, and when the brick begins its descent past the top of the building once more. Since we are finding the time the brick takes to travel in the vertical direction, only the y-component of the velocity is relevant, $\displaystyle v_{y} = (10\frac{m}{s})sin(30\degree)=5\frac{m}{s}$.

To find the time the brick is in the air for the first part, use the kinematic equation $\displaystyle v_y = v_{y,0}+a_{y}t$. The time the brick takes to reach its peak is given by  $\displaystyle 0 = 5\frac{m}{s} - (9.81\frac{m}{s^2})t$, so $\displaystyle t=0.51s$. Doubling this $\displaystyle t$ gives the time the brick takes to rise then fall again to the height from which it was thrown, so $\displaystyle t_{1} = 2t = 1.02s$.

To find the time the brick takes to fall from the top of the building to the ground, use the kinematic equation $\displaystyle \Delta y = v_{y,0}t+\frac{1}{2}a_{y}t^2$. We know that the initial velocity for this part, which is now pointing downward, is equal in magnitude to when the brick was first thrown, $\displaystyle v_{y,0} = -5\frac{m}{s}$. The height of the building is given by $\displaystyle \Delta y = -100m$. Hence our equation becomes $\displaystyle -100m = (-5 \frac{m}{s})t + \frac{1}{2}(-9.81 \frac{m}{s^2})t^2$. Solving for $\displaystyle t$ gives $\displaystyle t_{2}=4.03s$.

Adding the two times together, $\displaystyle t_{final} = t_{1} + t_{2} = 1.02s + 4.03s = 5.05s$.

In projectile motion, there are no horizontal forces acting on the object during its flight. By Newton's Second Law we know that if no force is applied to an object, it does not accelerate.

So, the cannonball's horizontal velocity remains constant. Twice the time of flight leads to twice the horizontal distance.

Determining the time for the upward velocity to equal zero:

$\displaystyle v_i+at=v_f$

$\displaystyle 40*sin60^\circ-9.8*t=0$

$\displaystyle t=3.53s$

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.

$\displaystyle 2t*v_x=d_x$

$\displaystyle 3.53*2*40cos 60^\circ=d_x$

$\displaystyle 141.2m$