Question is in the pic with options.
Please answer ASAP WITH EXPLANATION.
1ST ANSWER WILL BECOME BRAINLIEST...
HURRY !
Please answer ASAP WITH EXPLANATION.
1ST ANSWER WILL BECOME BRAINLIEST...
HURRY !
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Here is answer... (A) ...
As it is thrown upward let with velocity u m/s ....
At max height H .. the final velocity v = 0 ...
Let t be time .....
... u = gt .... (1) by first eqn .. v = u + at ....
As .. S = ut + 1/2 at^2 ...
H = ut - 1/2 g t^2 ..... (2) ....
put ... (1) in (2)...
we get... g = 2H / t^2 ..... (3)....
Now... let v be the velocity at height h ...
v^2 = u^2 - 2gh .... Third eqn ...of motion...
v^2 = (gt)^2 - 2 × 2H/t^2 × h .....from (3) & (1)...
v^2 = 4 / t^2 ( H^2 - Hh ) ...
v = 2 / t ( H^2 - Hh )^0.5 ....
put t = T /2 as T is total time of flight .. upward and then downward...
So ... v = 4/T ( H^2 - Hh )^0.5 .....
Hope u got it.....