Question

plzzz tell as soon as possible

Answer 1

Answer  3) 40 m/s.

    A particle is thrown upwards with a velocity u and it raises to a height h.   Let it take n seconds in its total flight. So it takes n/2 seconds upwards and n/2 sec. on its way down. Distance covered in n/2 th second is 35 m, before hitting the ground.

 Distance S_t  covered during the m th second is given by the formula:
      S_m = U + (m - 1/2) a, 
           here  U = 0,  a = g  = 10 m/s^2.   m = n/2.
      35 =  0 + (n/2 - 1/2) 10
          n = 8 sec.

n/2 = 4 sec. the particle takes to reach maximum height.

   So  apply v = u + a t   for the flight upwards.
                   0 = u - 10 * 4
                 u = 40 m/s.

Answer 2

Explanation:

Option 2 is correct answer !!