## Monday, March 27, 2023

### ChatGPT and Projectile Motion

In my previous post, I mentioned an exercise that I often give to my General Physics students that are learning about projectile motion. Using a PhET application, they found out that two different scenarios produced a projectile motion that has the same maximum height and same time of flight. I asked the students to figure out why.

I decided to post the same question to ChatGPT. The result was a bit of an eye-opener. I will show my question (in red) and its response below:

Why does a projectile shot with initial velocity of 20 m/s and angle 30 degrees with respect to the horizontal reach the same height as the projectile shot with initial velocity of 10 m/s and angle 90 degrees?

The key to understanding this phenomenon is to recognize that the vertical motion of the projectile is independent of its horizontal motion.
For both projectiles, the initial vertical velocity is zero. The only force acting on the projectile in the vertical direction is gravity, which causes the projectile to accelerate downward at a rate of 9.8 m/s^2. The vertical displacement of the projectile can be determined using the formula:
Δy = v0y*t + (1/2)at^2
where Δy is the vertical displacement, v0y is the initial vertical velocity, a is the acceleration due to gravity, and t is the time elapsed.
Since the projectile that was shot with an initial velocity of 10 m/s at an angle of 90 degrees goes straight up, it reaches its maximum height when its vertical velocity becomes zero. At this point, the time elapsed is given by:
t = v0y/a
where v0y is the initial vertical velocity. Substituting in the values, we get:
t = (10 m/s) / (9.8 m/s^2) = 1.02 s
Using this value of t in the above formula, we get:
Δy = (0 m/s)(1.02 s) + (1/2)(9.8 m/s^2)*(1.02 s)^2 = 5.1 m
So, the projectile reaches a maximum height of 5.1 m before falling back down to the ground.
For the projectile that was shot with an initial velocity of 20 m/s at an angle of 30 degrees, we can break down its initial velocity into its horizontal and vertical components:
v0x = v0cos(30) = 17.3 m/s
v0y = v0sin(30) = 10 m/s
The horizontal motion of the projectile is uniform, meaning it moves with a constant velocity of 17.3 m/s. The time taken for the projectile to reach its maximum height is the same as the time taken for it to reach half the range of its motion (i.e., the distance it travels horizontally before hitting the ground). The formula for the time of flight is:
t = 2*v0y/a
Substituting in the values, we get:
t = (2*10 m/s) / (9.8 m/s^2) = 2.04 s
The horizontal distance traveled by the projectile can be calculated using the formula:
R = v0x*t
Substituting in the values, we get:
R = (17.3 m/s)*(2.04 s) = 35.3 m
The maximum height reached by the projectile can be determined using the formula for vertical displacement:
Δy = v0y*t + (1/2)at^2
Substituting in the values, we get:
Δy = (10 m/s)(2.04 s) + (1/2)(9.8 m/s^2)*(2.04 s)^2 = 5.1 m
So, the projectile that was shot with an initial velocity of 20 m/s at an angle of 30 degrees reaches the same maximum height of 5.1 m as the projectile that was shot with an initial velocity of 10 m/s at an angle of 90 degrees, despite traveling much farther horizontally before hitting the ground.

Now, the answer is correct. In fact, it gave me more than what I asked for, i.e. it included the actual calculation of even the range, which I did not ask for.

However, its starting point is incorrect, and in fact, it is self-contradictory. It starts of by saying that the initial vertical velocity for both projectile is ZERO. This is clearly wrong.

Yet, a few lines down, it then mentioned and even found the vertical component of the initial vertical velocity, all while it was oblivious to what it just said in the beginning. So while the answer in the end may be correct, the reasoning, if we were to go by what it said, is puzzling and inconsistent.

I decided to challenge it by pointing out its mistake:

You are mistaken about the initial vertical velocity being zero. For both projectiles, the initial vertical velocity is not zero.

You are correct, I apologize for the mistake. The initial vertical velocity for the projectile shot with an initial velocity of 20 m/s and angle 30 degrees with respect to the horizontal is:
$v_{0y} = v_0 \sin \theta = 20 \text{ m/s} \sin 30^\circ = 10 \text{ m/s}$
And the initial vertical velocity for the projectile shot with an initial velocity of 10 m/s and angle 90 degrees with respect to the horizontal is:
$v_{0y} = v_0 \sin \theta = 10 \text{ m/s} \sin 90^\circ = 10 \text{ m/s}$
So, both projectiles have an initial vertical velocity of 10 m/s. The rest of the explanation and calculations remain the same.

This is now correct. Unfortunately, it requires someone who actually understands projectile motion to spot this. A student learning about it may not notice the inconsistency of the reasoning.

This seems to be the most common issue that I've come across and read about as far as scientific questions being asked to ChatGPT. This was similarly observed in another post that I wrote about. ChatGPT response has correct parts, but there are also parts that are puzzling, contradictory, and often downright wrong, all within one response.

There's more. In my next blog on ChatGPT, I asked it about buoyant forces. You don't want to miss that!

Zz.