Answer:
Speed of ball just before it hit the surface is 31.62 m/s .
Explanation:
Given :
Mass of ball , m = 50 g = 0.05 kg .
Height from which it falls , h = 80 m .
Thermal energy , E = 15 J .
Now , Initial energy of the system is :
[tex]E_i=\dfrac{mv^2}{2}+mgh[/tex]
Here , initial velocity is zero .
Therefore , [tex]E_i=mgh=40\ J[/tex]
Now , final energy of the system :
[tex]E_f=\dfrac{mv^2}{2}+mg(0)+15\\\\E_f=\dfrac{0.05\times v^2}{2}+15[/tex]
Since , no external force is applied .
Therefore , total energy of the system will be constant .
By conservation of energy :
[tex]E_i=E_f\\40=\dfrac{0.05v^2}{2}+15\\\\25=\dfrac{0.05v^2}{2}\\\\v=31.62\ m/s[/tex]
Therefore , speed of ball just before it hit the surface is 31.62 m/s .
