Acceleration=9.81m/s^2
initial velocity=0m/s
time=.28s
We have to find final velocity.
The equation we use is
Final velocity=initial velocity+acceleration x time
Vf=0m/s+(9.81m/s^2)(.28s)
Vf=2.7468m/s
We would round this to:
Vf (final velocity)=2.7m/s
If you dropped a ball from any height, and measured its distance from the ground at any regular interval while it's falling, the graph of that distance versus time would be a graph that curves downward.
-- The ball is falling down. As time goes on, it gets closer and closer to the ground. Its remaining distance from the ground keeps decreasing, so the line on the graph slopes down.
-- The speed of the ball keeps increasing (it accelerates) because of the gravitational force on it. As time goes on, it covers more of the remaining distance during each interval than it did in the previous interval. The downward slope of the graph keeps increasing.
Use the formula:
E = (1/2) * m * v ^2
For unit consistency, make sure that:
E energy is in joules (J)
m mass is in kilograms (kg)
v velocity is in meters per second (m/s)
So, substituting to the equation:
14= (1/2) * (17) * v^2
By algebraic manipulation, we derive:
v = (2*14/17)^0.5
v = 1.2834 m/s
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