Question

a ball is projected upwards with a velocity of 200m/s. find the speed of ball at the half of maximum height. take g= 10m/s^2

Answer 1

Assume the mass to be 'M'.
Kinetic energy on launch [tex]= \frac{1}{2}Mv^2 = \frac{1}{2}m(200)^2 = 20000M. [/tex]
This will transform into potential energy as the object rises up
At the highest point all kinetic energy will change into potential energy.
So, Potential energy at highest point = 20000M.
Potential energy is proportional to height.
So Potential energy at half maximum hight is half the Potential energy at highest point 
Potential energy at half maximum high = $\frac{20000M}{2}=10000M$.
 The rest will be kinetic energy which will be 10000M.
If v is the speed at half the maximum height,

then kinetic energy is $\frac{1}{2}Mv^2 = 10000M$

$Or \frac{1}{2}v^2 = 10000$

$v^2 = 20000$.

then speed, v = $\sqrt{20000}$

$v = 100\sqrt{2}\:ms^{-1}$