The aim here is to understand and explain the parabolic motion of a thrown object, known as projectile motion. Motion in one direction is unrelated to motion in other perpendicular directions. Once the object has been thrown, the only acceleration is in the \begin{align*}y\end{align*} (up/down) direction due to gravity. The \begin{align*}x\end{align*} (right/left) direction velocity remains unchanged.

Guidance

- In projectile motion, the horizontal displacement of an object from its starting point is called its
*range.*

- Vertical (\begin{align*} y \end{align*}) speed is zero only at the highest point of a thrown object's flight.

- Since in the absence of air resistance there is no acceleration in the horizontal direction, this component of velocity does not change over time. This is a counter-intuitive notion for many. (Air resistance will cause velocity to decrease slightly or significantly depending on the object. But this factor is ignored for the time being.)

- Motion in the vertical direction must include the acceleration due to gravity, and therefore the velocity in the vertical direction changes over time.

- The shape of the path of an object undergoing projectile motion in two dimensions is a parabola.

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### Simulations

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- Determine which of the following is in projectile motion. Remember that “projectile motion” means that gravity is the only means of acceleration for the object.
- A jet airplane during takeoff.
- A baseball during a Barry Bonds home run.
- A spacecraft just after all the rockets turn off in Earth orbit.
- A basketball thrown towards a basket.
- A bullet shot out of a gun.
- An inter-continental ballistic missile.
- A package dropped out of an airplane as it ascends upward with constant speed.

- Decide if each of the statements below is True or False. Then, explain your reasoning.
- At a projectile’s highest point, its velocity is zero.
- At a projectile’s highest point, its acceleration is zero.
- The rate of change of the \begin{align*}x\end{align*} position is changing with time along the projectile path.
- The rate of change of the \begin{align*}y\end{align*} position is changing with time along the projectile path.
- Suppose that after \begin{align*}2 \;\mathrm{s}\end{align*}, an object has traveled \begin{align*}2 \;\mathrm{m}\end{align*} in the horizontal direction. If the object is in projectile motion, it must travel \begin{align*}2 \;\mathrm{m}\end{align*} in the vertical direction as well.
- Suppose a hunter fires his gun. Suppose as well that as the bullet flies out horizontally and undergoes projectile motion, the shell for the bullet falls directly downward. Then, the shell hits the ground before the bullet.

- Imagine the path of a soccer ball in projectile motion. Which of the following is true at the highest point in its flight?
- \begin{align*} v_x = 0, v_y = 0, a_x =0, a_y = 0\end{align*}.
- \begin{align*} v_x > 0, v_y = 0, a_x =0, a_y = 0\end{align*}.
- \begin{align*} v_x = 0, v_y = 0, a_x =0, a_y = -9.8\;\mathrm{m/s}^2\end{align*}.
- \begin{align*} v_x > 0, v_y = 0, a_x =0, a_y = -9.8\;\mathrm{m/s}^2\end{align*}.

- A hunter with an air blaster gun is preparing to shoot at a monkey hanging from a tree. He is pointing his gun directly at the monkey. The monkey’s got to think quickly! What is the monkey’s best chance to avoid being smacked by the rubber ball?
- The monkey should stay right where he is: the bullet will pass beneath him due to gravity.
- The monkey should let go when the hunter fires. Since the gun is pointing right at him, he can avoid getting hit by falling to the ground.
- The monkey should stay right where he is: the bullet will sail above him since its vertical velocity increases by \begin{align*}9.8 \;\mathrm{m/s}\end{align*} every second of flight.
- The monkey should let go when the hunter fires. He will fall faster than the bullet due to his greater mass, and it will fly over his head.

- You are riding your bike in a straight line with a speed of \begin{align*}10 \;\mathrm{m/s}\end{align*}. You accidentally drop your calculator out of your backpack from a height of \begin{align*}2.0 \;\mathrm{m}\end{align*} above the ground. When it hits the ground, where is the calculator in relation to the position of your backpack? (Neglect air resistance.)
- You and your backpack are \begin{align*}6.3\;\mathrm{m}\end{align*} ahead of the calculator.
- You and your backpack are directly above the calculator.
- You and your backpack are \begin{align*}6.3 \;\mathrm{m}\end{align*} behind the calculator.
- None of the above.