Willing to pay 15fg per answers, i know its not much but it is still something! Please explain how you get the answers so then i will know how to do them.


Item #1

In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/s2 . Ignore air resistance.

How long is the flea in the air from the time it jumps to the time it hits the ground?
Express your answer in seconds to three significant figures.

Time Air = ?

Item #4
A test rocket containing a probe to determine the composition of the upper atmosphere is fired vertically upward from an initial position at ground level. During the time t while its fuel lasts, the rocket ascends with a constant upward acceleration of magnitude 2g. Assume that the rocket travels to a small enough height that the Earth’s gravitational force can be considered constant.

Part A:
What are the speed and height, in terms of g and t, when the rocket’s fuel runs out?
Express your answer in terms of g and t. V=


Part B:
Express your answer in terms of g and t. H=

Part C:
What is the maximum height the rocket reaches in terms of g and t?
Express your answer in terms of g and t. Hmax=

Part D:
If t = 34.0s , calculate the rocket’s maximum height. Hmax=


Hint: Item #4
A test rocket containing a probe to determine the composition of the upper atmosphere is fired vertically upward from an initial position at ground level. During the time t while its fuel lasts, the rocket ascends with a constant upward acceleration of magnitude 2g. Assume that the rocket travels to a small enough height that the Earth’s gravitational force can be considered constant.

(Part C) What is the maximum height the rocket reaches in terms of

g and t?



Hint: Break it into two sub-parts. (a) From 0 to t, the rocket accelerates from rest with acceleration plus 2g (plus for upward thrust from the engines). Find x and v at t. [A similar question would be: A sports car accelerates from rest with acceleration 3.0 m/s^2. After 10 seconds, where is the car?/what is the car's velocity?] (b) From t to maximum height, the 2g-engines shut off and the accelaration becomes gravity alone, or -g. Use the v just obtained, this is your new v_0 and your new a is minus g (minus for the downward force of gravity). It is asking for an x. Finally, the new v is 0 (because v = 0 means maximum height). This new x represents how much further the rocket rises after its 2g-engines shut off. Add this x to the x from subpart (a), and that's the answer.

(Part D) If t = 40.0s , calculate the rocket’s maximum height.

Substitute t = <your number> and g = 9.81 m/s^2 into the correct answer to part C.

Item #8

Learning Goal:
To apply the law of conservation of energy to an object launched upward in Earth's gravitational field.
In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy.
In this problem, you will apply the law of conservation of energy to different objects launched from Earth. The energy transformations that take place involve the object's kinetic energy K=(1/2)mv2 and its gravitational potential energy U=mgh. The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation
Ki+Ui=Kf+Uf,
where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem.

Part B
At the top point of the flight, what can be said about the projectile's kinetic and potential energy?

Both kinetic and potential energy are at their maximum values.
Both kinetic and potential energy are at their minimum values.
Kinetic energy is at a maximum; potential energy is at a minimum.
Kinetic energy is at a minimum; potential energy is at a maximum.

just need item #1 n 8

does number 1 just want the formula to calculate it in variables? because you gave no numbers...

it's physically impossible to know how long something is in the air for unless we know the time it was going up or falling... the height....


okay so you pmed me the velocity since it doesn't say it in the problem and you want time it takes to fall. Assuming you're correct and velocity is 2.66 m/s

then use kinematics eq

xfinal = 0 (because you land back at ground) xinitial= 0 (because you start at ground)
Your initial velocity is 2.66m/s.
gravity is 9.8m/s^2

xf = xi + vi(t) + 1/2(a)t^2.

0 = 0 + 2.62(t) + 1/2(-9.81)t^2
0 = 2.66t - 4.905t^2
4.905t = 2.66
t = 2.66/4.905
t = 0.542

This post was edited by DearJohnIllSeeYouSoon on Nov 5 2013 09:42pm

At the top point of the flight, what can be said about the projectile's kinetic and potential energy?
Both kinetic and potential energy are at their maximum values.
Both kinetic and potential energy are at their minimum values.
Kinetic energy is at a maximum; potential energy is at a minimum.
Kinetic energy is at a minimum; potential energy is at a maximum

kinetic energy is at a minimum, potential energy is at a maximum

object at highest point has 0 velocity in the Y direction, therefore it's at a minimum kinetic energy.

potential energy is at a maximum because acceleration is greatest at this moment because there is no accelerations going against each other.
For example: when a ball is going up... gravity pulls it down... but at the highest point it's no longer going up so it only has the potential energy of going down. there at a maximum

done thanks.