CBSE Class 11-science Answered
two balls a and b are released from rest from the roof of a building at t=0 and t=2s respectively.the ball a strikes the ground and rebounds back with the same speed.after some time both balls a and meet each other at a height of 55m from the ground.the height of the building is
Asked by Prasanna | 07 May, 2018, 09:17: AM
Expert Answer
Figure given above describes this question.
Let t1 be the time travelled by the ball, when it is freely falling through the height h of the building.
At end of t1 sec velocity v of the ball is given by v = g×t1 ..................(1)
The equation relating h and t1 is given by, h = (1/2)×g×t12 ..................(2)
ball-a rebounce and rise upto the height of 55 m just before meeting the ball-b. Let t2 be the time taken by the ball for this upward motion.
then we have 55 = v×t2-(1/2)×g×t22 ..................(3)
substitute v from (1) in (3), we have 55 = g×t1×t2 - (1/2)×g×t22 ..................(4)
Let ball-b falls through a height (h-55) in t3 seconds. Then we have
(h-55) = (1/2)×g×t32.....................................(5)
since ball-b dropped after 2 s, we have t1+t2 = t3+2 .................(6)
we have four unknowns, i.e., t1, t2, t3 and h. Hence we have to solve four equations i.e., (2), (4), (5) and (6) to get h
let us take g = 10 m/s ( lot of equatios to be solved. so let us avoid fractions)
let us rewrite eqn.(1) and (2) ; v = 10×t1 ..........................(1.a)
h = 5×t12 ............................(2.a)
rewrite eqn.(4) 11=2×t1×t2 - t22 ..................(3.a)
rewrite eqn.(5) h-55 = 5×t32 = 5×(t1+t2-2)2 = 5t12 + 5t22 + 20 +10×t1×t2 - 20×t1 - 20×t2 ...............(5.a)
substituting 5t12 from(2.a) and t22 from (3.a) in eqn.(5.a), cancelling equal terms beween LHS and RHS and after simplification we should get the following relation
t1t2+1-t1-t2 =0;
t2(t1-1)-(t1-1) = (t1-1)(t2-1) = 0; t2 < t1 ( t1 is the time taken to travel the height of the building downwards, whereas t2 is the time taken to travel only 55 m)
hence we get t2 = 1; if we substitute t2=1 in (3.a) we get t1 = 6; substitute t1 =6 in (2.a) we get h = 180 m
Answered by Thiyagarajan K | 07 May, 2018, 06:27: PM
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