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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 | 18 Jun, 2019, 13:47: PM
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 Thiyagarajan K | 18 Jun, 2019, 14:32: PM
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