a stone is dropped from the top of a 45m high building how fast it be moving when it reaches the ground and what will be the velocity be?​

30m/s

Explanation:

the distance(s) covered by a body in a straight line is expressed in terms of initial velocity(u),accelaration(a),and time(t)

now since the body is dropping free by the influence of gravity,(a) is a +10ms-²

therefore S=ut +½at²

Now the stone "dropped from the top".ie.from rest,so it doesn't have an initial velocity, (since it was not pushed)

hence u=0

the velocity of the body in a straight line is

V=u+at

so we have to find t in the first equation

and put it in the second one to find V, which is the velocity.

then S=½at² since u is 0

45=½•10•t²,after solving, we get t =3s

V=at since u is 0

V =10•3 =30m/s.

i hope this helps , you are free to ask any futhure questions.

the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.

no. dot product   =0 when vecs at right angles