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otez555 [7]
3 years ago
14

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?
Engineering
1 answer:
Dafna1 [17]3 years ago
7 0

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

Explanation :

By the 3rd equation of motion,

v^2-u^2=2as

where,

s = distance covered by the object = 6.93 m

u = initial velocity  = 2.99 m/s

v = final velocity = ?

a = acceleration = 9.8m/s^2

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

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

v=12.03m/s

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

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