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
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?
5 answers
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
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
1.35 m/s
8.0
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
-9.5=1/2(-9.81)t^2
t=0.44 s
x=vit
.352=vi(.44)
vi=0.8 m/s