IIT JEE , AIEEE Forum - Physics , Mechanics - answer this too simple question on kinematics and rates sure too!!!

krithika.r

A stone is dropped from a height of 45m. what will be the distance travelled by it during the last second of it's motion?

answer with detailed solution and explanation.

RATES FOR SURE!!

allamraju

Ans is 25m.Time period of motion=rt2H/g=rt2(45)/10=3s

So,the distance thro' it fell in first 2s=1/2(10)(2)²=20m and hence,distance travelled in last second=25m

GoNik

TRY USING THE NTH SECOND FORMULAE.........WILL SEE TO IT.......

chinmay_saxena01

the time taken by stone to reach the ground is =1/2gt^2 = 3sec.
now distance travelled by it in 3-1 sec. =20m.
so distance travelled by it in last sec of its motion is =45-20 = 25 m
allamraju is correct

d_utkarsh

first find the time taken to reach the ground by the formula S=ut-1/2*g*t*t

time comes to be 3 sec

taking t=2 sec find the distance covered by the ball

it comes out to be 20 mts

hence the distance covered in the last sec is 45-20=25 mt

raltz

since the stone is dropped initial vel. is 0.

then, h = 1/2gt² where t =3s.

using the formula,

S_n= u+g/2(2n-1) we get,

S_n= 10/2(5) = 25m

cheers!!!

karthik2007

@ Kritika - In case you are curious to know how we get the formula for the distance travelled in the nth second, this is how we proceed (I think you are in 10th, so in case you don't know, you could learn it)

Distance travelled in the nth second = Distance travelled in n seconds - Distance travelled in (n-1) seconds (think over this one deeply if you don't get it) = $[un + \frac{an^2}{2}] - [u(n-1) + \frac{a(n-1)^2}{2}] = u + \frac{a}{2}(2n-1)$ .

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Ans is 25m.Time period of motion=rt2H/g=rt2(45)/10=3s

So,the distance thro' it fell in first 2s=1/2(10)(2)2=20m and hence,distance travelled in last second=25m

So,the distance thro' it fell in first 2s=1/2(10)(2)²=20m and hence,distance travelled in last second=25m