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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
Answered by Expert
Answer:
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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