Apply equations of motion for free fall
Scenario
Aapne ball upar hawa mein phenki aur wo ek max height tak jakar ruk gayi, phir wapas aayi.
Kitni height tak gayi, aur usse neeche aane mein kitna time laga? Yeh sab hum bina stopwatch aur inch-tape ke kaise nikal sakte hain?
Mental model
Motion ke equations mein bas normal acceleration `a` ko `g` se replace kar do, baaki sab same rahega.
Kyunki free fall ek straight line uniform acceleration (constant `g`) ka case hai, isliye Chapter 8 ke purane teeno equations (`v = u + at`, `s = ut + 1/2 at^2`, `v^2 = u^2 + 2as`) yahan bhi perfectly apply hote hain. Bas `a` ban jayega `g`, aur `s` (distance) ban jayegi `h` (height).
Explanation
Motion ke `3` main equations hain. Jab object free fall kar raha hota hai, toh earth ke aas-paas `g` (`9.8 m/s^2`) lagbhag constant rehta hai. Isliye free fall, uniformly accelerated motion ka ek perfect example hai.
Hamein in equations ko use karte waqt sirf do baaton ka dhyan rakhna hota hai. Pehli: initial velocity (`u`) aur final velocity (`v`) kya hongi. Agar koi cheez upar se drop ki jaati hai (chhodi jaati hai), toh start mein rest par hoti hai, isliye `u = 0`. Agar cheez upar phenki jaati hai, toh sabse upar jakar ek fraction of second ke liye rukti hai, isliye final point par `v = 0`.
Doosri baat sign convention ki. Jo direction aapne positive li hai, uske opposite negative leni hogi. Agar motion downward direction mein hai, toh hum downward direction ko positive le sakte hain, aur `g` `+9.8 m/s^2` hoga. Agar aap ball upar phenk rahe hain, aur motion upward hai, toh acceleration (`g`) object ko rokne ki koshish kar raha hai, isliye `g` ko `-9.8 m/s^2` liya jayega.
Equations kuch aise dikhte hain free fall ke case mein:
v = u + gt
h = ut + 1/2 gt^2
v^2 = u^2 + 2gh
Jahan bhi `a` tha wahan `g`, aur `s` ki jagah `h` le aya gaya hai.
Worked example
Problem: A stone is released from the top of a tower of height `19.6 m`. Calculate its final velocity just before touching the ground. (Take `g = 9.8 m/s^2`)
- Kyunki stone 'released' ya drop kiya gaya hai, starting mein woh ruka hua tha.
- Hamein final velocity `v` nikalni hai aur time (`t`) nahi diya hua. Isliye hum third equation use karenge: `v^2 = u^2 + 2gh`.
- Multiply karte hain. `2 * 9.8` bhi `19.6` hota hai.
- Iska square root lekar `v` nikal lenge.
Answer: `19.6 m/s`
Problem: A ball is thrown vertically upwards with a velocity of `49 m/s`. Calculate the maximum height to which it rises. (Take `g = -9.8 m/s^2`)
- Upar ki taraf phenka gaya hai, toh shuruwati speed di gayi hai. Sabse upar jakar ball ruk jayegi.
- Time yahan bhi missing hai, to wapas `v^2 = u^2 + 2gh` lagayenge aur `h` ke liye solve karenge.
- Negative term ko dusri side bhejenge.
- `h` nikalne ke liye `19.6` se divide karenge.
Answer: `122.5 m`
Key points
- Dropped Object: Jab koi cheez chodi jaye (dropped), toh hamesha initial velocity `u = 0` lo.
- Object Thrown Upwards: At maximum height, final velocity becomes zero. During upward motion, `g` is taken as `-9.8 m/s^2` as it opposes the motion.
Common mistakes
- Ignoring the sign of g: Jab cheez upar ja rahi hai tab bhi bache `g` ko `+9.8` le lete hain, jisse answer mein error aa jata hai. Hamesha upward motion retard (slow down) karta hai isliye `g` negative hota hai.
Recall questions
- What is the final velocity of an object at its maximum height when thrown vertically upwards?
- What is the acceleration of a freely falling object?
- If an object is thrown upwards, what is its velocity at the highest point?
Questions & answers
A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s^2, find the maximum height reached by the stone.
80 m.
Approach: Use v^2 - u^2 = 2gs. Here v = 0, u = 40, a = -10. So, -1600 = 2(-10)h => h = 1600/20 = 80 m.
A ball is dropped from a height of 20m. Find time to reach ground (g=10).
2 s
Approach: Use h = ut + 0.5gt². 20 = 0 + 5t² => t² = 4 => t = 2s.
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