Thu December 08, 2011 By:

When a body is thrown upward so that it takes t seconds to reach the highest point, then a) Distance travelled in(t)th sec. =distance travelled in (t+1)th sec =5m b) distance travelled in (t-1)th sec. = distance travelled in (t+2)th sec. = 15m c) Distance travelled in (t-r)th sec. = distance travelled in (t+r+1)th sec. = (2r+1)5 m kindly explain these three one has derived these results.

Expert Reply
Fri December 09, 2011
distance travelled in t th sec=u+a(2t-1)/2=u-5(2t-1)
a=-10 m/s2
dist travelled by t+1 th sec=u-5(2t+1)
here dist travelled by t th  sec is the dist in t sec - dist travelled in (t-1) sec.
and dist in consequitive secs are in ratio 1:3:5 on
this comes from simple solving of eq of motions.
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