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CBSE Class 10 Answered

A body is dropped from a certain heigh H. If the ratio of distance travelled by it in (n-3)seconds to (n-3)nd secondsecond is 4:3, find H.(g = 10 m/s^2)  
Asked by acv27joy | 12 May, 2018, 08:18: PM
answered-by-expert Expert Answer
Let me say about what I understand about this question.
 
S1 is the distance travelled for (n-3) seconds from a height H after its dropping.
S2 is the distance travelled at (n-3)th second. i.e during the interval of (n-4) second to (n-3) second
Given ratio S1:S2 is 4:3 ; n seconds is the total time travel to cover the distance H.
 
Distance travelled upto (n-3) seconds,  S1 = (1/2)×g×(n-3)2   ......................(1)
speed after (n-4) second is , u = (n-4)×g ....................(2)
Distance travelled from (n-4)th second to (n-3)th second, S2 = u×1+(1/2)g×1×1 = (n-4)×g+(1/2)×g.........(3)
 
S1/S2 = [ (1/2)×g×(n-3)2] / [ (n-4)×g+(1/2)×g ] = 4/3 ........................(4)
 
solving the eqn.(4) for n, we get n=5;
 
Hence distance travelled H = (1/2)×g×5×5 = 125 m [ put g = 10 m/s2 ]
Answered by Thiyagarajan K | 13 May, 2018, 09:48: AM
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