How Long Does a Flea Stay in the Air After Jumping? Find Out Now!

[jnimagine]

A flea jumps straight up to 0.390m. How long does it stay in the air from the time it jumps up to the time it lands?

First of all, I found the initial velocity.
with the initial velocity, i tried using the equation d = v1t + 1/2at^2 to get the time.
but apparently it's wrong...;;

How do you figure this out?

 

[LowlyPion]

A flea jumps straight up to 0.390m. How long does it stay in the air from the time it jumps up to the time it lands?

First of all, I found the initial velocity.
with the initial velocity, i tried using the equation d = v1t + 1/2at^2 to get the time.
but apparently it's wrong...;;

How do you figure this out?

How long to fall from that height?

Double it, because it must have taken just as long to get up there.

 

[jnimagine]

How long to fall from that height?

Double it, because it must have taken just as long to get up there.

so the distance I find from the above equation is just one way??
d = v1t +1/2at^2
v1 would be the initial speed I found and once I find t using quadratic equation, if I double it, then it would be the total time it was in the air for going up and then coming back down?

 

[LowlyPion]

so the distance I find from the above equation is just one way??
d = v1t +1/2at^2
v1 would be the initial speed I found and once I find t using quadratic equation, if I double it, then it would be the total time it was in the air for going up and then coming back down?

Why bother with the velocity when the height yields time directly?

 

[jnimagine]

Why bother with the velocity when the height yields time directly?

so... which equation do i use if I don't use v1??

 

[LowlyPion]

so... which equation do i use if I don't use v1??

Which equation relates distance and time with constant acceleration?

 

[jnimagine]

Which equation relates distance and time with constant acceleration?

d, t, and a...
the only thing i can think of is d = v1t + 1/2at^2...T.T
is it wrong to use this equation and double it??
hmm... i can't think of another equation...:(

 

[jnimagine]

the thing is I already found v1 which was asked in the previous part...
but i don't know if using d+v1t + 1/2at^2 and doubling the answer makes sense...:|

 

[LowlyPion]

the thing is I already found v1 which was asked in the previous part...
but i don't know if using d+v1t + 1/2at^2 and doubling the answer makes sense...:|

It's a uniform gravitational field. That means that motion will have symmetry. That at every point on the way up, the object will have the same velocity magnitude, albeit reversed in direction, on the way down.

Since you know at apogee that it has velocity of 0, then the time to drop is given by X = 1/2 a*t².

But since that is only half the trip ... double it.

Why?

Because you can.

Good luck.

 

[jnimagine]

It's a uniform gravitational field. That means that motion will have symmetry. That at every point on the way up, the object will have the same velocity magnitude, albeit reversed in direction, on the way down.

Since you know at apogee that it has velocity of 0, then the time to drop is given by X = 1/2 a*t².

But since that is only half the trip ... double it.

Why?

Because you can.

Good luck.

but doesn't it have a certain v1 when it's going up even though the v1 for coming down is 0?

 

[LowlyPion]

but doesn't it have a certain v1 when it's going up even though the v1 for coming down is 0?

Sure it has a V1 to start out. And incidentally, it has the same V1 speed when it reaches the ground again just different direction.

But its like a video in reverse. The up trip and the down trip will appear the same, forwards or backwards.

So ... trip time up = trip time down.