# Time that an object is in the air

1. Nov 30, 2005

### Kandy

In a question, it asks for the time that an object is in the air; at first I thought it meant the time at max height, but I was told it should be the time at max height divided by 2 because the object goes up and down. But if the object goes up and down, wouldn't you mult by 2?

Last edited: Nov 30, 2005
2. Nov 30, 2005

### moo5003

The time it takes to reach max height x 2 would give you the total time it was in the air.

3. Nov 30, 2005

### Kandy

Here's the question:
An object fired straight upward with an initial velocity of 1.0 x 10^2 m/s is subjected to the effects of gravitation.
In a previous question, I solved for max height, which was 5.1 x 10^2 m. (I'm not sure if this is related to the next question though).
The time the object is in the air is
a. 4.5 s
b. 9.0 s
c. 20 s
d. 10 s

I keep on getting d. 10s

vi(velocity initial)= 100m/s vf(final)= 0 a(acceleration)= -9.81m/s^2 displacement=not needed t(time)= ?

a = (vf - vi) / t
t = (vf - vi) / a
t = (0 - 100m/s) / (-9.81m/s^2)
t = (-100m/s) / (-9.81m/s^2)
t = 10s but the correct ans should be a. 4.5s

Last edited: Nov 30, 2005
4. Nov 30, 2005

### kp

hmm...sounds strange too me. 4.5s total time for a projectile at shot 100m/s?

5. Nov 30, 2005

### Kandy

well, i could be wrong about total time, the question maybe asking for smthg else... I geuss I'll check if the ans key is wrong tomorrow...

6. Nov 30, 2005

### kp

yeah I get H = 510.20 meters, total time 20.4seconds, time at H(max) = 10.2 seconds

7. Nov 30, 2005

### Kandy

and even if you divide by 2 for whatever reason, you still wouldn't get 4.5s, so...

well, thanks!