Question

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

Answer if know the answer and can explain it
physics lovers are welcomed​

Answer 1

Answer:

(A) t1 + t2 = 2u/g

Explanation:

When a body, that is thrown vertically with velocity u,  reaches a vertical distance h from ground at time t1,

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

If H is maximum the height it reaches with time T, then we will 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 the vertical distance travelled during ascending will be ( H-h ). Thus -  

(H - h) = (1/2) gt² ---(4)

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

If t2 is the time for the object, when it starts from the ground, and reaches maximum height and  again crosses the spot which is at a distance h from ground,  then we have,  

t = (t2 - T )  --- (5)

Writing Eqn.(4) 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 simplifying equation (6), we will get,

t1 + t2 = 2u/g