Question

a stone is thrown vertically up with an initial velocity 49 M per second from the top of a tower and reaches the ground after 12 second find the height of the tower

Answer 1

The stone is thrown up with certain velocity U (as in the write - up) from the top of a tower of height say h.

naturally it will go up to some height h' depending on its velocity but due to pull of earth it has to stop and turn back...falling backward it will accelerate and will reach the tower height h with velocity of same magnitude U but speeding down.

therefore its final velocity v will depend on h

one can use relation between initial and final velocities as function of h and g

v^2 - U^2 = 2.g.h

as v is given as 3.u(small u) ,

9.u*2 - U^2 = 2.g. h therefore one can say h = (1/2g) { 9u^2}

now i doubt if capital U was a typo error ...in case its U= u then

h= (8u^2)/ 2.g

Answer 2

d=s/t
d=49×12
=588
this will help you