TBR height approximation for projectiles

[bluv1212]

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This is an explanation for the sample projectile problem on page 32 in the TBR physics I book. I posted on an older post but didn't get any responses since last night and I wanna move on now, but not knowing this makes me feel like i'm not making the most of my money .

"From that, we can approximate that the apricot climbs about 40 m from launch to
its apex (based on the approximation that a 1-s climb leads to a height of 5 m, a 2-
s climb leads to a height of 20m, and a 4-s climb leads to a height of 80 m). This
means that at its highest point, the apricot is roughly 350 m above the ground."


I've seen this explanation in dealing with the turbo-method solution and projectiles in a PDF posted previously by BerkReviewTeach. can anyone can clarify how we arrive at this approximation and how to use it - is there a formula, its like exponential right? I just want to be able to know how to use it if it comes up. It seems like its necessary in order to use the turbo-method solution, but i'm not sure because its not explicitly explained.If anyone can explain this to me, i would really appreciate it.

Thanks so much.

 

[Will Hunting]

This is an explanation for the sample projectile problem on page 32 in the TBR physics I book. I posted on an older post but didn't get any responses since last night and I wanna move on now, but not knowing this makes me feel like i'm not making the most of my money .

"From that, we can approximate that the apricot climbs about 40 m from launch to
its apex (based on the approximation that a 1-s climb leads to a height of 5 m, a 2-
s climb leads to a height of 20m, and a 4-s climb leads to a height of 80 m). This
means that at its highest point, the apricot is roughly 350 m above the ground."


I've seen this explanation in dealing with the turbo-method solution and projectiles in a PDF posted previously by BerkReviewTeach. can anyone can clarify how we arrive at this approximation and how to use it - is there a formula, its like exponential right? I just want to be able to know how to use it if it comes up. It seems like its necessary in order to use the turbo-method solution, but i'm not sure because its not explicitly explained.If anyone can explain this to me, i would really appreciate it.

Thanks so much.

Ok, it takes anything generally 1 second to fall 5meters. It takes it 2 seconds to fall 20 meters. It takes it 3 seconds to 45 meters. 4 seconds to all 80.

How you do it is to take the Distance (40) /5 and take the square root of that to get how long. For example, in one of their examples they have an apricot fall off 310 meters. How long? Well 310/5=62 so take square root of 65=8 seconds. The actual is a little less than 8.

In your example it's 40/5=8. The square root of 8 is between 2 and 3 and closer to 3 seconds.

so it's (D/5)^.5 Any time you raise something to .5 it means you're taking a square root.

 

[ThirdEye]

This is an explanation for the sample projectile problem on page 32 in the TBR physics I book. I posted on an older post but didn't get any responses since last night and I wanna move on now, but not knowing this makes me feel like i'm not making the most of my money .

"From that, we can approximate that the apricot climbs about 40 m from launch to
its apex (based on the approximation that a 1-s climb leads to a height of 5 m, a 2-
s climb leads to a height of 20m, and a 4-s climb leads to a height of 80 m). This
means that at its highest point, the apricot is roughly 350 m above the ground."


I've seen this explanation in dealing with the turbo-method solution and projectiles in a PDF posted previously by BerkReviewTeach. can anyone can clarify how we arrive at this approximation and how to use it - is there a formula, its like exponential right? I just want to be able to know how to use it if it comes up. It seems like its necessary in order to use the turbo-method solution, but i'm not sure because its not explicitly explained.If anyone can explain this to me, i would really appreciate it.

Thanks so much.

g = 10m/s^2 which means that after every second the velocity will either increase or decrease by 10m/s, depending on direction.

10m/s^2 x 1s = 10m/s
10m/s^2 x 2s = 20m/s
10m/s^2 x 4s = 40m/s


Now to find the height it goes (either up or down, it's all the same) in X seconds, all you do is take the AVERAGE velocity and multiply it by the time.

For 1 second, V = 10m/s^2 x 1s = 10m/s ---> Average velocity = (V1 +v2)/2 = 5m/s ----> 5m/s x 1s = 5m

For 2 seconds, V = 10m/s^2 x 2s = 20m/s ---> Avg V = (20+0)/2 = 10m/s ---> 10m/s x 2s = 20m

For 4 seconds, V = 10m/s^2 x 4s = 40m/s----> avg v = 20m/s ---->20m/s x 4s = 80m


Audio Osmosis is a really excellent tool for helping build intuition for translational motion.

 

[bluv1212]

thanks guys. This clears up a lot