A ball is dropped off a very tall canyon ledge. Gravity accelerates the ball at 9.8 m/s2. How fast is the ball traveling after 5 seconds?

Respuesta :

As ball is dropped, so its initial velocity (u) of the ball is = 0

Acceleration of the ball (a) is given to be 9.8 m/s^2

As acceleration is constant, so we can apply the equation of kinematics for constant acceleration i.e. [tex]v = u + at[/tex]

where u is initial velocity, a is acceleration and v is final velocity at after time t

so velocity after 5 sec will be [tex]= 0 +(9.8)5 = 49 m/s[/tex]