Physics, asked by iitaspirant52, 11 months ago

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​


iitaspirant52: plz tell

Answers

Answered by abhi178
4

answer : option (A) t1 + t2 = 2u/g

explanation : a body is thrown vertically with a velocity u. it passes a point B at a height of h above the ground at t1 time . while going up at point C and at time t2 while falling down at A.

hence, time taken to pass at point B = t1

and time taken to move from point B to C and then to fall down at point A = t2

here, t1 + t2 = time of flight .....(1)

Let time taken to reach maximum height H by body is T.

at maximum height velocity of body is 0

from formula, v = u + at

v= 0, t = T and a = -g

so, T = u/g......(2)

time of flight = time of ascending+ time of descending

from equation (1)

or, T + T = t1 + t2 [ motion under gravity, time of ascending= time of descending]

or, 2T = t1 + t2

from equation (2),

t1 + t2 = 2u/g

Attachments:
Answered by Anonymous
1

Answer:

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

Similar questions