Question

A stone falls from rest. The distance covered by it in the last second of its motion is equal to the distance covered
in the first three seconds. What is the height from which the stone was dropped? [Take g = 10 m/s]
(1) 25 m
(2) 100 m
(3) 125 m
(4) 200 m
​

Answer 1

Answer:

ans is in pic

Explanation:

hope it helps

Answer 2

here u=0; g=10m/s

from question that distance cover in last second is equal to distance cover in first three second

♠️so we can said that Sⁿ=S ..............(1) and time for first part of motion is 3 second

S=ut+1/2 at²

S=0t+10*3*3/2

S=90/3

S=45m

since from equation (1)

Sⁿ=u+1/2a(2n-1)=45

0+10(2n-1)=90

20n-10=90

20n=100

2n=10

n=5

mean total journey of partical is of 5 second

♠️since we have distance covered by partical in first three second is 45m and for last second mean 5th second is also 45m (according to question).

we have to find motion during 4th second, which is given by

Sⁿ=u+1/2a(2n-1)

S⁴=0+1/2*10*(2*4-1)

S⁴=0.5*10*7

S⁴=0.5*70

S⁴=35m

♠️since total distance covered is equal to sum of all three motion which is equal to height from stone was dropped

=S³th+S⁴th+S⁵th

=45+35+45

=125m

♠️since hight is equal to 125 m

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