v = at = -9.8(.333) = -3.26
s = 6.06 - 4.9t^2
when it hits, s=0, so
0 = 6.06 - 4.9t^2
t = 1.11
An object falls from the top of the school, which is 6.06m high. Assume air resistance can be ignored and the acceleration is 9.8m/s^2
If the object was dropped( not thrown) , what is its speed after 0.333s
How long does it take to hit the ground
2 answers
The speed at t = 0.333 s is
V = (g/2) t^2
It hits the ground when (g/2) t^2 = 6.06 m
Solve for that new t.
V = (g/2) t^2
It hits the ground when (g/2) t^2 = 6.06 m
Solve for that new t.
1 answer