Horizontally launched projectile | Two-dimensional motion | Physics | Khan Academy

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Summary

This video from Khan Academy explains how to solve problems involving horizontally launched projectiles. It highlights a common mistake students make and clarifies the independence of horizontal and vertical motion.

Highlights

What is a horizontally launched projectile?
00:00:18

A horizontally launched projectile is any object starting with a purely horizontal velocity. Examples include a ball rolling off a table or a person running straight off a cliff, meaning there's no initial vertical velocity.

Example Problem: Cliff Diver
00:01:00

The video presents an example of a cliff diver running off a 30-meter cliff at 5 meters per second. The goal is to determine the horizontal distance covered before hitting the water.

Independent Motion: Horizontal vs. Vertical
00:02:31

A key concept is that horizontal and vertical motions are independent. The horizontal velocity (5 m/s) remains constant throughout the flight, assuming no air resistance or propulsion. The vertical velocity, however, starts at zero and increases downwards due to gravity.

Setting up the Equations and Common Pitfall
00:04:05

The video outlines the known variables for both horizontal and vertical directions. A crucial point is that the vertical displacement is -30 meters (since the diver falls downward). The most common mistake is assuming the initial vertical velocity is 5 m/s; it is actually 0 m/s because the launch is purely horizontal.

Solving for Time
00:07:51

To solve for the horizontal displacement, first calculate the total time of flight using the vertical motion equations. Using the kinematic equation delta y = v_iy * t + 0.5 * a_y * t^2, and plugging in the values (delta y = -30m, v_iy = 0, a_y = -9.8 m/s^2), the time is found to be 2.47 seconds.

Solving for Horizontal Displacement
00:10:02

Once the time is known, it can be used in the horizontal motion equation (delta x = v_ix * t + 0.5 * a_x * t^2). Since horizontal acceleration (a_x) is zero, the equation simplifies to delta x = v_ix * t. Plugging in 5 m/s for v_ix and 2.47 seconds for t, the horizontal displacement (dx) is calculated to be approximately 12.4 meters.

Key Takeaways
00:12:02

The main points to remember for horizontally launched projectile problems are to use negative displacement for downward motion and, most importantly, to recognize that the initial vertical velocity is zero, even if not explicitly stated.

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