Projectile Motion Maximum Height Equation, Physics | PART 2

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This lecture is about deriving the maximum height of the projectile motion and how to calculate the maximum height of the projectile motion.

Projectile Motion Playlist:

Introducation to Projectile Motion:

Time of Flight in Projectile Motion:

Q: What is the maximum height of the projectile motion?

Ans: The highest vertical position of an object from the ground from where it falls down is called maximum height of projectile motion.
For example, when you launch a ball at an angle, the highest vertical position of the ball from the ground is called its maximum.

Finally, I will teach you the exam based numerical of maximum height of the projectile motion with solution.

To learn more about the maximum height pf the projectile motion, watch this animated lecture till the end.

#ProjectileMotionHeight
#MaximuHeightt
#physics



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Untold truth
This is the most underrated channel I have ever seen😕
U deserve more🥰

iit
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You have an unique talent to explain the most difficult topic in an easiest way.♥️🌿

harshita
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U deserve more . I understood only by seeing this once.

tpsg
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the formula is (50sin30) ^2/ 2(9.8)
for those who are lost

jaydenclarke
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Wow it is an interesting explanation and keep it up ❤️❤️

hermelagkidan
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Beautiful explication sir, iam really flatted, tqq

jayajayalakshmi
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From Bangladeshi🙂...it was very vwry helpul for me🙂❤️

shakibsarkar
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God bless you I’ve really understood bless you

blessedtugz
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Good evening, I have a doubt.Can u tell me why u filled the value of g as positive when the projectile is in the upwards direction??

jiwanjyoti
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Sir formula is h = u sq sin sq theta divided by 2 times of g. Thankyou for explaining too simply 🎉❤

JEE-G
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I hate physics but this channel make me love it

jedenny
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Sir plz tell the imp questions in physics sir

peyyilasubbulakshmi
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Hi could you please explain about range in projectile motion

sultanrashid
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Help me!!!
Why is acceleration due to Gravity positive when it is thrown vertically upward!!!?

ilangmeie
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Why we ignore the velociyy in x direction ?

amnaaslam
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There is a mistake in the equation
H=u²sin²theta/2g

darkfury
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I keep getting 0 whenever I do the final calculations? Explain how you got 31.9?

pillowcase
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I think y'all should try this equation instead to get 31.9 m

H max = 1/2 u² sinø / 2g

kkfxqgh
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This is not a lecture 😂 it's a short video😅

Smile_world