Question

35. A particle is projected vertically upwards from a
point A on the ground. It takes time ti to reach a
point B, but it still continues to move up. If it takes
further t2 time to reach the ground from point B.
Then height of point B from the ground is
(a) = 9 (t + t2? (b) g t t2.
(6) 5g (6 + t2}
(
agutz​

Answer 1

Let the particle is thrown with speed u from a point A on the ground. after time t1, particle reaches a point B. but it still continues to move up and it takes t2 time to reach the ground from point B.

so, time of flight = (t1 + t2)

then, time taken to reach maximum height = (t1 + t2)/2

initial velocity , u = g(t1 + t2)/2 [ as you know, formula v = u + at , here v = 0 , a = -g and t = (t1 + t2)/2 ]

now, the height of point B from the ground is S = ut1 - 1/2 gt1²

= g(t1 + t2)/2 × t1 - 1/2 gt1²

= 1/2 g(t1.t2 + t1² ) - 1/2gt1²

= 1/2 gt1.t2

hence, height of point B from the ground is 1/2gt1.t2