mass is tied to spring and begins periodically . the distance between its highest and its lowest position is 48cm. what is the a
1 answer:
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1) Time in the air: 0.78 s
The motion of the ball is a projectile motion, which consists of two independent motions:
- A horizontal motion with constant horizontal velocity
- A vertical motion with constant downward acceleration of
(acceleration of gravity)
The initial vertical velocity is
where the negative sign means the direction is downward.
The vertical position of the ball is given by
where
h = 4.50 m is the initial heigth of the ball when it starts falling down
The ball reaches the ground when y = 0, so we have:
This is a second-order equation; solving for t, we get
t = -1.18 s
t = 0.78 s
We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.
2) Horizontal distance: 1.83 m
For this second part of the problem, we just have to consider the horizontal motion of the ball.
As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by
where u = 3.05 m/s is the initial speed and the angle of projection.
For a uniform motion, we can use the following relationship between distance covered and velocity:
and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground:
Answer:
1) The maximum jump height is reached at A.
2) The maximum center of mass height off of the ground is B.
3) The time of flight is C.
4) The distance of jump is B.
Explanation:
First of all we need to decompose velocity in its rectangular components, so
1) We use, , as we clear it for
and using the fact that
at max height, we obtain
2) We can use the formula for
, so
3) We can use the formula , to find total time of fligth, so
, as it is a second-grade polynomial, we find that its positive root is
4) Finally, we use , as it has an additional displacement of
due the leg extension we obtain,
, aprox