(1)
Cheetah speed:
Its position at time t is given by
(1)
Gazelle speed:
the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by
(2)
The cheetah reaches the gazelle when . Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:
(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.
Cheetah:
Gazelle:
So, the gazelle should be ahead of the cheetah of at least
Answer:
h = 3.3 m (Look at the explanation below, please)
Explanation:
This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy
Kinetic energy = m
Plug in the numbers = (4.0)(
)
Solve = 2(64) = 128 J
Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.
Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)
Kinetic energy = Potential Energy
128 J = 39.2h
h = 3.26 m
h= 3.3 m (because of significant figures)
Answer:
Approximately .
Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.
Explanation:
Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant near the surface of the earth.
For an object that is accelerating constantly,
,
where
In this case, is the same as the change in the ball's height:
. By assumption, this ball was dropped with no initial velocity. As a result,
. Since the ball is accelerating due to gravity,
.
.
In this case, would be the velocity of the ball just before it hits the ground. Solve for
.
.