Jim Guinn’s PHYS1111 Assignment #3 Solutions
1.
A ball is thrown vertically (you don’t know up or down) out of a window in a tall building.
It takes 3.30sec to pass a point 20 meters below where it started.
a.)
What was the ball’s initial velocity?
Choosing our origin at the lower position of the ball, and upward
as positive, we have the values given on the left.
We can then solve
x
o
= 20m
x = 0
v
o
= ?
x = x
o
+ v
o
t + (1/2)a
o
t
2
or with x = 0 we have
v =
v
o
= [x
o
– (1/2)a
o
t
2
] / t = [20m – (1/2) (9.80m/s
2
) (3.3s)
2
] / (3.3s) =
a
o
= 9.80m/s
2
v
o
= 10m/s
t = 3.30s
The ball was thrown upward at 10m/s .
b.)
What is the velocity of the ball when it passes the point 20 meters below?
Since we don’t have any equations that yield a final velocity without using the initial
velocity, we’ll use
v = v
o
+ at = 10m/s + (9.80m/s
2
) (3.3s) =
v = 22m/s .
2. An object has a position as a function of time x(t) = 4m t
3
/s
3
+ 5m t/s .
a.)
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 Spring '11
 lomant
 Velocity

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