How to Solve Difficult Projectile Motion Problems (Angled Launch from a height)

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3 жыл бұрын

This video goes over examples of projectiles launched at angles from various heights or launched to some height. These are typically harder problems in physics.

Пікірлер: 11

@mamudouconateh8308 Жыл бұрын

Thank you so much

@HowToPhysics Жыл бұрын

Glad I could help!

@kendrasweetheart2320 11 ай бұрын

What is another way to answer part c

@medicisounds1384 5 ай бұрын

So i tried it a slight different way. One that i find easier I first Max Heigh Ball reaches using -- H= V^2 Sin^2(THETA) / 2g After i got my Max Height of 7.15769 meters I then used that to find TIME. t= sqrt(2(H + h) / g with this i got time of 1.2086 seconds After i used this formula to find RANGE R= Vx t Vx being V* Cos 37 My final answer was 14.4 m with roundings Im still learning physics! But i just figured it out this way and thought it was easier for me thank you

@medicisounds1384 5 ай бұрын

I know my answer was off by 2 meters but hey is physics ever exact ;) LOL

@HowToPhysics 5 ай бұрын

This is a totally acceptable way to solve the problem! When using sin and cos, error will propagate if you round the decimals so being off by 2m sounds reasonable.

@user-mb4or8ep8e Жыл бұрын

Sir , which equations are you using ? can you please make a video of the equations used so that we can understand.

@anzarv 11 ай бұрын

SUVAT.

@yin873 Жыл бұрын

for b, how do you know y = 0 rather than y-naught is equal to 0

@HowToPhysics Жыл бұрын

Great question! You can set your initial height to zero but if you do that your answer for the final height will end up being negative since it is below the initial height. That works but I prefer to use the ground or final position as zero so that when I find the initial height (y naught) I get a positive number. Either way works if you ignore the negative and just look at the magnitude of your answer. Hope this helps!

@kotlin7554 10 ай бұрын

thx@@HowToPhysics