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# A body is thrown vertically upwards from ground with a velocity u. It passes a point at a height h above the ground at time t1 while going up and at a time t2 while falling down. (t1 and t2 are measured from the instant of projecting body upwards). Then the relation between u, t1 and t2 and is:- (A) t1 + t2 = 2u/g (B) t2 - t1 = 2u/g (C) t1 + t2 = u/g (D) t2 - t1 = u/g

Asked by Kkrishna 18th June 2019, 1:47 PM
when a body, which is thrown vertically with velocity u,  reaches a vertical distance h from ground at time t1,

then we have, h = u×t1 - (1/2)g×t12  ............................(1)

if H is maximum height it reaches with time T, then we have

H = u2 / (2g)  ..........................(2)

T = u/g  ............................... (3)

During descending, if the object crosses the same spot which is at a height h above the ground,
then vertical distance travelled during ascending is ( H-h ) .  Then we have

(H - h) = (1/2) g t2 ......................(4)

where t is time taken to travel from highest point to the distance (H-h) .

If t2 is the time for the object, when it is starting from ground , reaches maximum height and
again crossing the spot which is at a distance h from ground,

then we have,   t = (t2 - T )  .......................(5)

Eqn.(4) is rewritten by substituting for H, h and t using respective eqns. (2), (1), (3) and (5)

[ u2 / (2g) ] - u×t1 + (1/2)g×t12 = (1/2)g [ t2 - (u/g) ]2 ...............................(6)

After simplification of eqn.(6), we get,  t1 + t2 = 2u/g
Answered by Expert 18th June 2019, 2:32 PM
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