How to Solve the Kinematics Ratio Problem for a Particle's Journey?

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SUMMARY

The discussion focuses on solving the kinematics ratio problem for a particle's journey, specifically determining the time ratio \( t1:t2 \) where \( t1 \) is the time taken to reach \( \frac{15}{16}H \) and \( t2 \) is the time taken to fall the remaining distance to the ground. The participant attempted to use equations of motion but struggled to derive the correct ratio from the options provided: 1:3, 3:1, 5:3, and 3:5. Key insights include recognizing that \( t2 \) must be greater than \( t1 \) and the need for additional formulas related to free fall and projectile motion to solve the problem accurately.

PREREQUISITES
  • Understanding of projectile motion and kinematics
  • Familiarity with equations of motion, specifically \( s = ut + \frac{1}{2}gt^2 \)
  • Knowledge of maximum height calculations in projectile motion
  • Basic concepts of free fall and gravitational acceleration
NEXT STEPS
  • Study the derivation of time of flight in projectile motion
  • Learn how to calculate time taken to fall from various heights using \( t = \sqrt{\frac{2h}{g}} \)
  • Explore the relationship between time and distance in free fall scenarios
  • Review advanced kinematic equations and their applications in solving complex motion problems
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators seeking to clarify concepts related to time ratios in motion problems.

croslali
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I was given this worksheet from my teacher and well I am finding it really tough so far .. the question below is just crushing my skull .. so i would appreciate it if you guys would guide me through it

"A particle projected from ground takes time 't1' to reach 15/16 H and time 't2' to cover the rest of distance to the ground. If 'H' is the maximum height attained. find the ratio t1: t2

well my options were many ... they were 1:3 ,3:1, 5:3 and 3:5

and well using my knowledge of atleast what i know is: 1)the object has almost reached the max height so the ratio won't be that big 2)i tried using the equations of motions and to no avail i still didnt get anywhere 3)[usin(thetha)(t1)]-[(1/2)g(t1)^2]=(15/16)H 4)and [(usin(thetha)^2]/2g=H

keeping in mind that t2 has to obviously be greater than t1 ... the second and the third options are eliminated ... well it surely can't be 1:3(maybe) .. because that is a really big ratio ...

is there some other extra formula or something other than the common formulae (like max height and range and splitting of vectors)i am supposed to know to solve this question am i missing out anything?
 
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Try this. How long does it take to fall from a height H? How long does it take to fall a distance H/16?
 

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