Help Me With Physics!

[Voski]

OK so all of my homework is online and to make things worst my teacher hasn't given us any formulas or done any examples so I have no idea what formulas to use to solve the problems. I am reading the book but I am pretty brain dead now.

Question 1 : "A flea jumps straight up to a maximum height of 0.380 . What is its initial velocity as it leaves the ground?"
Answer 1: I got 2.73 m/s

Question 2 : "How long is the flea in the air from the time it jumps to the time it hits the ground?"

So I can't figure out what to use for question 2. I am assuming it isn't that hard I just have no idea where to start.

[liquidglass]

I'm no physics expert by far. But using a logical approach I would say that you use your first answer to apply to the 2nd one. Figure out how many seconds he would travel to reach .380.

Basically my approach would be to figure out what percentage of 2.73m .380 is then multiple that by seconds. Up and down. Hope that makes sense.

[omgitsjimmy]

time = 2 * v / g

i'm tired, i'm going to bed i'll explain later Edited October 1, 2009 by omgitsjimmy

[K1024]

OK so all of my homework is online and to make things worst my teacher hasn't given us any formulas or done any examples so I have no idea what formulas to use to solve the problems. I am reading the book but I am pretty brain dead now.

Question 1 : "A flea jumps straight up to a maximum height of 0.380 . What is its initial velocity as it leaves the ground?"
Answer 1: I got 2.73 m/s

Question 2 : "How long is the flea in the air from the time it jumps to the time it hits the ground?"

So I can't figure out what to use for question 2. I am assuming it isn't that hard I just have no idea where to start.

question 1 : The formula you want to use here is V^2 = (Vo)^2 + 2*a*(x-xo)
that is Current Velocity Squared = Velocity(start) Squared + 2*acceleration*(current x - original x)
if you take the start velocity at the top of the jump (0) and the original x to be the top of the jump (.380) and the current x to be the ground (0) you can plug those in and find V. a is acceleration, which in this case is gravity (-9.8 assuming your using metric)
you get
V^2 = 0^2 + 2*(-9.8)*(0-.380)
V^2 = 7.4480
V = 2.73 m/s like you have.

question2 : the formula you need here is V = Vo + a*t
that is Velocity = Velocity(start) + acceleration * time again, you use Vo = 0, a = -9.8 and V = -2.73 (negative because its falling) and solve for t
-2.73 = 0 + (-9.8)*t
t = .278 s but this isnt the full answer, only the part of time from the top of the jump to the bottom. since acceleration is constant, we know that it took the same amount of time to go up as come down, so time is doubled and the answer is
t = .557 seconds

hope that helps

[Voski]

OK so all of my homework is online and to make things worst my teacher hasn't given us any formulas or done any examples so I have no idea what formulas to use to solve the problems. I am reading the book but I am pretty brain dead now.

Question 1 : "A flea jumps straight up to a maximum height of 0.380 . What is its initial velocity as it leaves the ground?"
Answer 1: I got 2.73 m/s

Question 2 : "How long is the flea in the air from the time it jumps to the time it hits the ground?"

So I can't figure out what to use for question 2. I am assuming it isn't that hard I just have no idea where to start.

question 1 : The formula you want to use here is V^2 = (Vo)^2 + 2*a*(x-xo)
that is Current Velocity Squared = Velocity(start) Squared + 2*acceleration*(current x - original x)
if you take the start velocity at the top of the jump (0) and the original x to be the top of the jump (.380) and the current x to be the ground (0) you can plug those in and find V. a is acceleration, which in this case is gravity (-9.8 assuming your using metric)
you get
V^2 = 0^2 + 2*(-9.8)*(0-.380)
V^2 = 7.4480
V = 2.73 m/s like you have.

question2 : the formula you need here is V = Vo + a*t
that is Velocity = Velocity(start) + acceleration * time again, you use Vo = 0, a = -9.8 and V = -2.73 (negative because its falling) and solve for t
-2.73 = 0 + (-9.8)*t
t = .278 s but this isnt the full answer, only the part of time from the top of the jump to the bottom. since acceleration is constant, we know that it took the same amount of time to go up as come down, so time is doubled and the answer is
t = .557 seconds

hope that helps

Oh I see. For the second part isn't Vo = 2.73? and V=0? Since we solved for Vo in the first part of the question. Either way it is the same answer because you used -2.73.

Thanks for the help!

[Voski]

OK I have one more I don't know how to start this one. I was messing around but I don't think I am doing it right.

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 50.0 , and the distance between them is 53.0 . After = 2.00 , the motorcycle starts to accelerate at a rate of 8.00 . The motorcycle catches up with the car at some time .

1. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2-t1.

2. How far does the motorcycle travel from the moment it starts to accelerate (at time t1) until it catches up with the car (at time t2)?

[Tyler]

If the limit never approaches 0 then the limit doesn't exist!

[Stuie]

Because Schrödinger hated cats... and Future Quantum Physics Students.