A 3.52 kg steel ball is tossed upward from a height of 6.93 meters above the floor with a vertical velocity of 2.99 m/s. What is the velocity (m/s) of the ball just before it strikes the floor?

Respuesta :

Answer : The final velocity of the ball is, 12.03 m/s

Explanation :

By the 3rd equation of motion,

[tex]v^2-u^2=2as[/tex]

where,

s = distance covered by the object = 6.93 m

u = initial velocity  = 2.99 m/s

v = final velocity = ?

a = acceleration = [tex]9.8m/s^2[/tex]

Now put all the given values in the above equation, we get the final velocity of the ball.

[tex]v^2-(2.99m/s)^2=2\times (9.8m/s^2)\times (6.93m)[/tex]

[tex]v=12.03m/s[/tex]

Thus, the final velocity of the ball is, 12.03 m/s