A student drops a ball from the top of a 37-meter tall building. The ball leaves the student's hand with a zero speed. What is the speed of the ball at the moment just before it hits the ground?

Respuesta :

Answer:

The ball accelerates because of gravity which is constant. So we can apply one of the equations of uniformly accelerated motion to solve the problem.

[tex]Vf^2=Vo^2+2aS\\[/tex] is compatible since [tex]S, Vo, a[/tex] are given enabling us to calculate [tex]Vf[/tex] .

[tex]Vf^2=(0m/s)^2+2*10m/s^2*37m\\Vf^2=20m/s^2*37m\\Vf^2=740m^2/s^2\\Vf=\sqrt{740} m/s\\ Vf=27.202941017 m/s\\[/tex]