It is necessary to understand how to break a vector into its x and y components in order to solve problems for projectiles.

#### Break the Initial Velocity Vector into its Components

#### Apply the Kinematics Equations

\begin{align*} \textbf{Vertical Direction} && \textbf{Horizontal Direction}\\ y(t) = y_i +v_{iy} t - \tfrac{1}{2} g t^2 && x(t) = x_{i} +v_{ix}t\\ v_{y}(t) = v_{iy} -gt && v_{x}(t) = v_{ix} \\ {v_y}^2 = {v_{0y}}^2 - 2g (\Delta y) && \\ a_y = -g = -9.8 \mathrm{m/s^2} \approx -10\mathrm{m/s^2} && a_x = 0\\ \end{align*}

To work these problems, separate the “Big Three” equations into two sets: one for the vertical direction, and one for the horizontal. Keep them separate.

The only variable that can go into both sets of equations is time; use time to communicate between the x and y components of the object's motion.

### Examples

#### Example 1

CSI discovers a car at the bottom of a 72 m cliff. How fast was the car going if it landed 22m horizontally from the cliff’s edge? (Note that the cliff is flat, i.e. the car came off the cliff horizontally).

\begin{align*}v = \ ? \ [m/s]\end{align*}

Given: \begin{align*}h = \Delta y = 72 \ m\end{align*}

\begin{align*}{\;}\qquad \quad d = \Delta x = 22 \ m\end{align*}

\begin{align*}{\;}\qquad \quad g = 10.0 \ m/s^2\end{align*}

Equation: \begin{align*}h = v_{iy} t + \frac{1}{2} gt^2\end{align*} and \begin{align*}d = v_{ix} t\end{align*}

Plug n’ Chug:

Step 1: Calculate the time required for the car to freefall from a height of 72 m.

\begin{align*}h = v_{iy} t + \frac{1}{2} gt^2\end{align*} but since \begin{align*}v_{iy}=0\end{align*}, the equation simplifies to \begin{align*}h = \frac{1}{2} gt^2\end{align*} rearranging for the unknown variable, \begin{align*}t\end{align*}, yields

\begin{align*}t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(72 \ m)}{10.0 \ m/s^2}} = 3.79 \ s\end{align*}

Step 2: Solve for initial velocity:

\begin{align*}v_{ix} = \frac{d}{t} = \frac{22 \ m}{3.79 \ s} = 5.80 \ m/s\end{align*}

The answer is 5.80 m/s.

#### Example 2

A ball of mass \begin{align*}m\end{align*} is moving horizontally with a speed of \begin{align*}v_i\end{align*} off a cliff of height \begin{align*}h\end{align*}. How much time does it take the ball to travel from the edge of the cliff to the ground? Express your answer in terms of \begin{align*}g\end{align*} (acceleration due to gravity) and \begin{align*}h\end{align*} (height of the cliff).

Since we are solving or how long it takes for the ball to reach ground, any motion in the \begin{align*}x\end{align*} direction is not pertinent. To make this problem a little simpler, we will define down as the positive direction and the top of the cliff to be:\begin{align*}y=0\end{align*}In this solution we will use the equation:\begin{align*}y(t)=y_o+v_{oy}t+\frac{1}{2}gt^2\end{align*}

\begin{align*} y(t)&=y_o+v_{oy}t+\frac{1}{2}gt^2 && \text{start with the equation}\\ h&=y_o+v_{oy}t+\frac{1}{2}gt^2 && \text{substitute } h \text{ for } y(t) \text{ because that's the position of the ball}\\ & && \text{when it hits the ground after time } t\\ h&=0+v_{oy}+\frac{1}{2}gt^2 && \text{substitute } 0 \text{ for } y_o \text{ because the ball starts at the top of the cliff}\\ h&=0+0+\frac{1}{2}gt^2 && \text{substitute } 0 \text{ for } v_{oy} \text{ because the ball starts with}\\ & && \text{no vertical component to its velocity}\\ h&=\frac{1}{2}gt^2 && \text{simplify the equation}\\ t&=\sqrt{\frac{2h}{g}} && \text{solve for } t\\ \end{align*}

### Interactive Simulation

### Review

- A stone is thrown horizontally at a speed of \begin{align*}8.0 \;\mathrm{m/s}\end{align*} from the edge of a cliff \begin{align*}80 \;\mathrm{m}\end{align*} in height. How far from the base of the cliff will the stone strike the ground?
- A toy truck moves off the edge of a table that is \begin{align*}1.25 \;\mathrm{m}\end{align*} high and lands \begin{align*}0.40 \;\mathrm{m}\end{align*} from the base of the table.
- How much time passed between the moment the car left the table and the moment it hit the floor?
- What was the horizontal velocity of the car when it hit the ground?

- A hawk in level flight \begin{align*}135\;\mathrm{m}\end{align*} above the ground drops the fish it caught. If the hawk’s horizontal speed is \begin{align*}20.0 \;\mathrm{m/s}\end{align*}, how far ahead of the drop point will the fish land?
- A pistol is fired horizontally toward a target \begin{align*}120\;\mathrm{m}\end{align*} away, but at the same height. The bullet’s velocity is \begin{align*}200 \;\mathrm{m/s}\end{align*}. How long does it take the bullet to get to the target? How far below the target does the bullet hit?
- A bird, traveling at \begin{align*}20 \;\mathrm{m/s}\end{align*}, wants to hit a waiter \begin{align*}10 \;\mathrm{m}\end{align*} below with his dropping (see image). In order to hit the waiter, the bird must release his dropping some distance before he is directly overhead. What is this distance?
- Joe Nedney of the
*San Francisco 49ers*kicked a field goal with an initial velocity of \begin{align*}20 \;\mathrm{m/s}\end{align*} at an angle of \begin{align*}60^\circ\end{align*}.- How long is the ball in the air?
*Hint:*you may assume that the ball lands at same height as it starts at. - What are the range and maximum height of the ball?

- How long is the ball in the air?
- A racquetball thrown from the ground at an angle of \begin{align*}45^\circ\end{align*} and with a speed of \begin{align*}22.5 \;\mathrm{m/s}\end{align*} lands exactly \begin{align*}2.5 \;\mathrm{s}\end{align*} later on the top of a nearby building. Calculate the horizontal distance it traveled and the height of the building.
- Donovan McNabb throws a football. He throws it with an initial velocity of \begin{align*}30 \;\mathrm{m/s}\end{align*} at an angle of \begin{align*}25^\circ\end{align*}. How much time passes until the ball travels \begin{align*}35\;\mathrm{m}\end{align*} horizontally? What is the height of the ball after \begin{align*}0.5\end{align*} seconds? (Assume that, when thrown, the ball is \begin{align*}2 \;\mathrm{m}\end{align*} above the ground.)
- Pablo Sandoval throws a baseball with a horizontal component of velocity of \begin{align*}25 \;\mathrm{m/s}\end{align*}. After \begin{align*}2\end{align*} seconds, the ball is \begin{align*}40 \;\mathrm{m}\end{align*} above the release point. Calculate the horizontal distance it has traveled by this time, its initial vertical component of velocity, and its initial angle of projection. Also, is the ball on the way up or the way down at this moment in time?
- Barry Bonds hits a \begin{align*}125\;\mathrm{m}(450')\end{align*} home run that lands in the stands at an altitude \begin{align*}30 \;\mathrm{m}\end{align*} above its starting altitude. Assuming that the ball left the bat at an angle of \begin{align*}45^\circ\end{align*} from the horizontal, calculate how long the ball was in the air.
- A golfer can drive a ball with an initial speed of \begin{align*}40.0 \;\mathrm{m/s}\end{align*}. If the tee and the green are separated by \begin{align*}100\;\mathrm{m}\end{align*}, but are on the same level, at what angle should the ball be driven? (
*Hint:*you should use \begin{align*} 2\cos{(x)}\sin{(x)} = \sin{(2x)} \end{align*} at some point.) - How long will it take a bullet fired from a cliff at an initial velocity of \begin{align*}700 \;\mathrm{m/s}\end{align*}, at an angle \begin{align*}30^\circ\end{align*} below the horizontal, to reach the ground \begin{align*}200 \;\mathrm{m}\end{align*} below?
- A diver in Hawaii is jumping off a cliff \begin{align*}45 \;\mathrm{m}\end{align*} high, but she notices that there is an outcropping of rocks \begin{align*}7 \;\mathrm{m}\end{align*} out at the base. So, she must clear a horizontal distance of \begin{align*}7 \;\mathrm{m}\end{align*} during the dive in order to survive. Assuming the diver jumps horizontally, what is his/her minimum push-off speed?
- If Monte Ellis can jump \begin{align*}1.0 \;\mathrm{m}\end{align*} high on Earth, how high can he jump on the moon assuming same initial velocity that he had on Earth (where gravity is \begin{align*}1/6\end{align*} that of Earth’s gravity)?
- James Bond is trying to jump from a helicopter into a speeding Corvette to capture the bad guy. The car is going \begin{align*}30.0 \;\mathrm{m/s}\end{align*} and the helicopter is flying completely horizontally at \begin{align*}100 \;\mathrm{m/s}\end{align*}. The helicopter is \begin{align*}120\;\mathrm{m}\end{align*} above the car and \begin{align*}440 \;\mathrm{m}\end{align*} behind the car. How long must James Bond wait to jump in order to safely make it into the car?
- A field goal kicker lines up to kick a \begin{align*}44\end{align*} yard \begin{align*}(40 \;\mathrm{m})\end{align*} field goal. He kicks it with an initial velocity of \begin{align*}22 \;\mathrm{m/s}\end{align*} at an angle of \begin{align*}55^\circ\end{align*}. The field goal posts are \begin{align*}3\end{align*} meters high.
- Does he make the field goal?
- What is the ball’s velocity and direction of motion just as it reaches the field goal post (
*i.e.,*after it has traveled \begin{align*}40\;\mathrm{m}\end{align*} in the horizontal direction)?

- In a football game a punter kicks the ball a horizontal distance of \begin{align*}43\end{align*} yards \begin{align*}(39 \;\mathrm{m})\end{align*}. On TV, they track the hang time, which reads \begin{align*}3.9\end{align*} seconds. From this information, calculate the angle and speed at which the ball was kicked.
*(Note for non-football watchers: the projectile starts and lands at the same height. It goes \begin{align*}43\end{align*} yards horizontally in a time of \begin{align*}3.9\end{align*} seconds)*

### Review (Answers)

- \begin{align*}32 \;\mathrm{m}\end{align*}
- a. \begin{align*}0.5 \;\mathrm{s}\end{align*} b. \begin{align*}0.8 \;\mathrm{m/s}\end{align*}
- \begin{align*}104 \;\mathrm{m}\end{align*}
- \begin{align*}t = 0.60 \;\mathrm{s}, 1.8 \;\mathrm{m}\end{align*} below target
- \begin{align*}28 \;\mathrm{m}\end{align*}.
- a. \begin{align*}3.5 \;\mathrm{s}\end{align*}. b. \begin{align*}35 \;\mathrm{m}; 15 \;\mathrm{m}\end{align*}
- \begin{align*}40 \;\mathrm{m}; 8.5 \;\mathrm{m}\end{align*}
- \begin{align*}1.3\end{align*} seconds, \begin{align*}7.1\end{align*} meters
- \begin{align*}50 \;\mathrm{m}; v_{0y} = 30 \;\mathrm{m/s}; 50^0\end{align*}; on the way up
- \begin{align*}4.4 \;\mathrm{s}\end{align*}
- \begin{align*}19^\circ\end{align*}
- \begin{align*}0.5 \;\mathrm{s}\end{align*}
- \begin{align*}2.3 \;\mathrm{m/s}\end{align*}
- \begin{align*}6 \;\mathrm{m}\end{align*}
- \begin{align*}1.4\end{align*} seconds
- a. yes b. \begin{align*}14 \;\mathrm{m/s}\end{align*} @ \begin{align*}23\end{align*} degrees from horizontal
- \begin{align*}22 \;\mathrm{m/s}\end{align*} @ \begin{align*}62\end{align*} degrees