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CBSE Class 9 Answered

Suppose the object is dropped from the height of 5m. ,then at 1 sec. its accelaration will be 9.8m/s*s . Question is that will it be19.6 (9.8*2) at the end of 2 Sec. ?
Asked by Ranjit Kothari | 29 Oct, 2010, 12:00: AM
answered-by-expert Expert Answer

Dear Student,

In this case, acceleration is constant = g. At 1 second, it is indeed 9.8 m/s2;
Since s = ut + 1/2 gt2       
              =  1/2 gt2 (u=0)  and height = 5m,
object will hit the ground between t=1 and t=2 seconds. Hence its acceleration at t=2 will be 0. (assuming no bouncing)
 
If the height had been much greater than 5 m, acceleration at t=2 would still be = 9.8 m/s2. Velocity will be 9.8 m/s and 2*9.8 m/s respectively at t=1 and t=2 seconds in such a case.

We hope that clarifies your query.
Regards
Team
Topperlearning

Answered by | 30 Oct, 2010, 12:55: AM

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