A stone is thrown vertically upward with a speed of 27.0 m/s .How fast is it moving when it is at a height of 14.0 m? How much time is required to reach this height?

Question

mathmate · Answer

Since this is a post in physics, we'll use the usual kinematics equation.
v^2-u^2=2aS
v=final velocity
u=initial velocity
a=acceleration
S=distance
Substitute numbers,
v=14 m/s (going up)
u=27 m/s (going up)
a=-9.81 m/s^2 (going down)
S=(v^2-u^2)/2a=\(14^2-27^2)/(-9.81)=?
(The answer should be positive)
Note that v could be negative, i.e. v=-14 m (going down).
But the answers are the same, since v is squared.

For the second part, 
use 
v(t)=v(0)+at
where a=-9.81
v(t)=14 m/s
v(0)=27 m/s
solve for t.

"How much time is required to reach this height" implies time required while going UP, so v(t)=-14 m/s (going down) can be discarded.

As you can see, once the basic equations of kinematics have been mastered (or even just memorized), solutions follow.

sachintha · Answer

mathmate you have misinterpreted the question. 
For the first part it asks for the velocity at the height of 14m. So,
$v^2-u^2=2as$
$\,\qquad v_a = \sqrt{2as+u^2}$
$\:\,\quad\qquad = \sqrt{2\times-9.8\times14+27^2}$

For the second part,
$v=u+at$
$\,t = \Large\frac{v-u}{a}$
$\:\:\;\, = \Large\frac{v_a-27}{-9.8}$

Simplify to get the answers.

mathmate · Answer

Very true.  I missed that.  Thank you for point it out. 
@sachintha