Question

a boy on a cliff 19.6m high drops a stone. one second later, he throws a second stone after the first. they hit the ground at the same time. with what speed did he throw the second stone

Answer 1

Let T be the time taken by the 1st stone to travel 19.6 m.

s = u t + 1/2 a t²  for the 1st stone
   19.6   =  1/2 * g * T²
   T =  √(39.2/ 9.8) = 2 sec

2nd stone:
   19.6 = u * 1sec + 1/2 g 1²
    u = 14.7 m/sec
This is the speed with which the second stone is thrown.  

Answer 2

Answer:

Let T be the time taken by the 1st stone to travel 19.6 m.

s = u t + 1/2 a t²

 19.6   =  1/2 * g * T²  

T =  √(39.2/ 9.8)

    = 2 sec

so, This is the speed for the first stone thrown.  

now for the 2nd stone, we have to find U :

so, 19.6 = u * 1sec + 1/2 g 1²    

therefore,

u = 14.7 m/sec

so, This is the speed for the second stone thrown.  

pls mrk as brainliest