Question

A body of mass 5kg is thrown vertically up with a kinetic energy of 490 J. Find the height at which the kinetic energy of the body becomes half of the original value.

Answer 1

K.E=0.5m*$v^{2}$

490=5*0.5*$v^{2}$

√(490*2)/5=v

v=14m/s----(Initial velocity u)

Now substitute K.E as 245 (490/2) and find v which come out to be 7√2---(Final velocity v)

$v^{2}=u^{2}+2as$

From the above equation we get

$(7\sqrt{2})^{2}=14^{2}-2*9.8*s$(a=acceleration due to gravity which is in the opposite direction of initial velocity hence -ve)

Hence by solving the equation we get s=15m which is the answer