Question

plzzz solve the problem

Answer 1

Hey mate (Allenite) here is the correct answer.....

Time taken for the particle to reach highest point is t1+t2/2

(v=u−gt)

Therefore initial velocity=u=g(t1+t2/2)

Therefore heigh of B from ground is

h=ut+1/2at^2

h=ut1−1/2gt1^2

=g(t1+t2/2)t1−1/2gt1 ^2

h=1/2gt1t2

Hence d is the correct answer.

I hope it will be helpful to you....

Please mark me as BRAINLIEST ❤▶❤❤..

Answer 2

LET THE HIGHEST POINT WHERE THE BALL REACHES AFTER POINT B be C.

TIME TAKEN BY BALL TO REACH B IS t1 AND TIME TAKEN TO (REACH C FROM B+C to A) is =t2.

We know that,,,,

Time taken to reach max.height is = time taken to reach ground from that point.

Therefore ,Total time of flight=(t1+t2)

Time taken to reach C from A=(t1+t2)÷2

From A to C -; v=u+(-g)t 0=u-g (t1+t2)÷2u u=(gt1+gt2)÷2

using,

s=ut+0.5at^2

Now,

by putting value of u,

a=-g,t=t1

we get-

AB=½gt1t2

HOPE IT HELP...

☣AMAN☣