Answer:
3.07 m/s
Explanation:
We know that from kinematics equation
and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity
Making u the subject then

Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then

The answer is point b because vertical velocity is zero at the maximum height
)
5
-5
1 2 3
4
5
Other than at t = 0, when is the velocity of
the object equal to zero?
1. 5.0 s
2. 4.0 s
3. 3.5 s
4. At no other time on this graph. correct
5. During the interval from 1.0 s to 3.0 s.
Explanation:
Since vt =
Z t
0
a dt, vt
is the area between
the acceleration curve and the t axis during
the time period from 0 to t. If the area is above
the horizontal axis, it is positive; otherwise, it
is negative. In order for the velocity to be zero
at any given time t, there would have to be
equal amounts of positive and negative area
between 0 and t. According to the graph, this
condition is never satisfied.
005 (part 1 of 1) 0 points
Identify all of those graphs that represent motion
at constant speed (note the axes carefully).
a) t
x
b) t
v
c) t
a
d) t
v
e) t
a