Question

n a cliff 19.6 m high drops a stone. One second later, he throws a second stone after They both hit the ground at the same time. With what speed did he throw the second?

Answer 1

Given, h = 19.6m g = 9.8m/s², u = 0

1st Stone :

Formula is  h = ut + 1/2 at^2
                       = 19.6 = 0 + 9.8t²/2t
                       = 2 sec

Second Stone : 

Formula is h = ut + 1/2 at^2
                      = 19.6 = u*1+1/2u
                      = 14.7


Hope this helps!

Answer 2

Answer:

Explanation:

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