Question

14. A body dropped from the top of a tower covers a distance
7x in the last second of its journey, where x is the distance
covered in first second. How much time does it take to reach
the ground?
(a) 3s (6) 45 (6) 5s (d) ås​

Answer 1

WE need to use the formulas :- S = u×t + (1/2)g×t2 and v = u + gt
where u is initial speed, v is speed after t seconds, g is acceleration due to gravity and S is distance travelled after t seconds

Distance covered in first second = (1/2) g = x ...............(1)

( In eqn.(1), since the body is dropped from top of tower u = 0 also t = 1 s )

Let the time taken by the body to hit the ground be n seconds

velocity after (n-1) seconds = (n-1)g ...................(2)

distance covered in nth second = (n-1)g+(1/2)g = 7x ..................(3)

using eqn.(1) and eqn.(3), we have, (n-1)g+(1/2)g = (7/2)g or n = 4 seconds please mark as brainlyest

Answer 2

Ans=4 second

DATE:
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14) Sath - ut & (2n-1) using formula
oslo
Grote]
U=0
Enth = 7X
7x = otto (2n-1)
7
Angain weget value
TX = 5(20-1) I rom earmuw son
7x = (2n-1) rolet Si = x
S : Ut g (2n-1)
nBecause
Si = 01 tos (2x1-1)
n = + (x + 1) L
Put the value x S. 5 (2-1) value of
n = 1 ( 7x5+1) S: 5 L
i. Si = x 25
x =5
n = 4 sec And
Time to reach the ground.
Total