# A steel ball rolls with constant velocity across a tabletop 0.950m high. IT rolls off and hits the ground +0.352m horizontally from the edge of the table. How fast was the ball rolling?

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1. y=vit+1/2at^2
-9.5=1/2(-9.81)t^2
t=0.44 s

x=vit
.352=vi(.44)
vi=0.8 m/s

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2. 8.0

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3. Get the time to fall from the height, H = 0.950 m. Call that time T. You will need to use the acceleration of gravity (g) to compute T.
H = (g/2) T^2
Haven't you seen that equation before for falling bodies?

Then use the relationship
V T = 0.352
to get V

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4. 1.35 m/s

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5. A steel ball rolls with constant velocity across a tabletop 0.950m high. IT rolls off and hits the ground +0.352m horizontally from the edge of the table. How fast was the ball rolling?

GIVEN:

Vi = 0 m/s
d = 0.950 m ---> 0.352 m
a = 9.81 m/s^2
t = ?

d = Vit + 1/2 at^2

0.352 = 1/2 (9.82 m/s^2) (t)^2

0.352 = 4.905t^2

--> Divide by 4.095

square root - 0.071763507 = t^2

0.26 s = t

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