Introduction to Projectile Motion - Formulas and Equations

Main Ideas

Projectile motion is analyzed by separating it into horizontal ((x)) and vertical ((y)) motion. Both directions share the same elapsed time, but their accelerations differ:

• Horizontal: Acceleration is zero, so horizontal velocity (v_x) remains constant.
• Vertical: Gravity acts downward, with magnitude (g \approx 9.8\ \text{m/s}^2). Taking upward as positive, (a_y=-g).
• Air resistance is ignored.

Kinematics Equations

For constant velocity:

[ d=vt ]

For constant acceleration:

[ v_f=v_i+at ]

[ v_f^2=v_i^2+2ad ]

[ d=v_{\text{avg}}t=\frac{v_i+v_f}{2}t ]

[ d=v_it+\frac12at^2 ]

Here, (d) represents displacement. Distance and displacement are equal only when the object does not reverse direction.

Launching Horizontally from a Cliff

The object begins with horizontal velocity (v_x) and no initial vertical velocity.

The time to fall from height (H) is:

[ H=\frac12gt^2,\qquad t=\sqrt{\frac{2H}{g}} ]

The horizontal range is:

[ R=v_xt ]

Just before impact, (v_x) is unchanged and the vertical velocity is (v_y=-gt). The impact speed is found from the velocity components:

[ v=\sqrt{v_x^2+v_y^2} ]

The direction can be described by the angle below the horizontal:

[ \tan^{-1}\left(\frac{|v_y|}{v_x}\right) ]

Launching and Landing at the Same Height

For a launch speed (V) at an angle (\theta) above the horizontal, resolve the initial velocity into components:

[ v_x=V\cos\theta,\qquad v_y=V\sin\theta ]

When the projectile lands at its launch height, its trajectory is symmetric:

• Time to reach maximum height: [ t_{\text{up}}=\frac{V\sin\theta}{g} ]

• Total flight time: [ t_{\text{total}}=\frac{2V\sin\theta}{g} ]

• Maximum height above the launch point: [ H_{\max}=\frac{V^2\sin^2\theta}{2g} ]

• Horizontal range: [ R=\frac{V^2\sin(2\theta)}{g} ]

At the top of the path, the vertical velocity is zero. With no air resistance, the projectile’s impact speed equals its launch speed, and its impact direction is the launch angle reflected below the horizontal.

Launching from an Elevated Position at an Angle

If the projectile is launched from height (H), it does not land at its launch height. Therefore, the equal-height flight-time and range formulas do not apply directly.

Taking upward as positive, the vertical position is:

[ y(t)=H+V\sin\theta\,t-\frac12gt^2 ]

To find the time to hit the ground:

1. Set (y=0), the ground level.
2. Solve the resulting quadratic equation for (t).
3. Use the positive, physically meaningful time.

Alternatively, split the flight into two parts:

1. Find the time to the peak: [ t_{\text{up}}=\frac{V\sin\theta}{g} ]

2. Find the height gained above the launch point: [ H_{\max}=\frac{V^2\sin^2\theta}{2g} ]

3. The peak is (H+H_{\max}) above the ground. The time to fall from the peak is: [ t_{\text{down}}=\sqrt{\frac{2(H+H_{\max})}{g}} ]

4. Add the times, then calculate the range: [ R=v_xt_{\text{total}}=V\cos\theta\,t_{\text{total}} ]

At impact, calculate the vertical velocity using:

[ v_y=V\sin\theta-gt_{\text{total}} ]

Combine it with the unchanged horizontal velocity to find the impact speed and direction.

Key Reminders

• Choose the (x)- or (y)-direction equation that matches the quantity being solved.
• Keep the signs of vertical velocity and acceleration consistent with the chosen positive direction.
• The formula (R=V^2\sin(2\theta)/g) applies specifically when the projectile lands at the same height from which it was launched.
• For impact angles, distinguish a reference angle below the horizontal from an angle measured counterclockwise from the positive (x)-axis.

Speakers or Sources Featured

• The Organic Chemistry Tutor — the video’s narrator and instructor. No other speakers or sources are identified in the subtitles.