Answer:
6.50m/s
Explanation:
Using conservation of energy, the total energy in the system is conserved. We know that the ball has potential energy initially because it is elevated 2 meters. We also know it has kinetic energy because it is given an initial speed of 1.75 m/s. At the bottom of the ramp, it loses all potential energy and is converted directly to kinetic energy. We can model the equation as below. U is the potential energy, K is the initial kinetic energy and K' is final kinetic energy.
[tex]U + K = K'[/tex]
Equation for kinetic energy is
[tex]K = 0.5mv^2[/tex]
and potential energy is:
[tex]U=mgh[/tex]
Plug these equation to the first equation and simplify (v' is final speed):
[tex]mgh+0.5mv^2= 0.5mv'^2\\gh+0.5v^2=0.5v'^2\\2gh+v^2=v'^2\\\sqrt{2gh+v^2}=v'[/tex]
Plug in the values:
[tex]v'= \sqrt{2gh+v^2}=\sqrt{2(9.81)(2)+(1.75)^2} \\ v' = 6.50m/s[/tex]
Final speed is 6.50m/s
