2-dimensional projectile motion (part 3)

TL;DR

The video explains how to calculate the total time and horizontal distance traveled by a projectile launched at a 45-degree angle with a horizontal velocity of 10 m/s.

Transcript

Welcome back. So we just figured out how long does this thing stay in the air. So what can we do with that? Let me clear all of this and redraw it because maybe every time I redraw it, it might further clarify things. Let me actually draw things a little bit fancier this time. So I have the ball and it's moving-- look at that. Let me just draw the ... Read More

Key Insights

• 🛩️ The total time of flight for a projectile launched at 45 degrees with a horizontal velocity of 10 m/s is equal to 2 square root of 2 seconds.
• 👱 Assuming no air resistance, the horizontal velocity of a projectile remains constant throughout its motion.
• 🚥 The horizontal distance traveled by a projectile is determined by multiplying the average horizontal velocity by the time of flight.
• 😥 The maximum height reached by a projectile is calculated using the average vertical velocity and the time it takes to reach the highest point.

Questions & Answers

Q: How long does the projectile stay in the air?

The projectile stays in the air for 2 square root of 2 seconds, as it takes the same amount of time to go from the ground to its highest point and back down.

Q: Why can we assume the horizontal velocity of the projectile remains constant?

The horizontal velocity stays constant because there is no force acting horizontally on the projectile, and assuming no air resistance, it will continue moving at the same speed.

Q: What is the formula for calculating the horizontal distance traveled by the projectile?

The horizontal distance is calculated by multiplying the average velocity (5 square root of 2 m/s) by the time the projectile is in the air (2 square root of 2 seconds), resulting in 20 meters.

Q: How can we determine the maximum height reached by the projectile?

The maximum height is found by calculating the average vertical velocity (5 square root of 2 over 2 m/s) and multiplying it by the time taken to reach the highest point, which is equal to the time taken to reach the ground (square root of 2 seconds). Thus, the maximum height is 5 meters.

Summary & Key Takeaways

• The video demonstrates how to calculate the time a projectile takes to go from the ground to its highest point and back down, which is equal to 2 square root of 2 seconds.

• By assuming no air resistance and constant horizontal velocity, the video explains how to calculate the horizontal distance traveled by the projectile, which is 20 meters.

• The video also shows how to calculate the maximum height reached by the projectile, which is 5 meters.