How fast (metric units) is a ball going after 3.0 seconds if you drop it off a very high building? It would be 9.8 m/s^2 since gravity is a constant speed, right?
If there is no external force acting on a free falling body then it must fall with the speed of 9.8 m/s^2(acceleration due to gravity) irrespective of its mass or any physical quantity.
Acceleration due to gravity is always different on different planets. which depends on the density of the celestial body.
So the answer wouldn't be 29.4 m/sec^2?
or actually -29.4 m/sec^2
v=v0-gt v0 = 0 v = -gt v = -9.8 * 3 = -29.4 m/s ???
Since they just want how fast and not which direction, you can ignore the - sign, so just 29.4m/s
for a free fall \[u=0\] \[g=9.8 m/s ^{2}\] putting this value in v=u+gt we get \[v=0+9.8*3 \] (since time is 3 second) therefore v=29.4\[m/s\]
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