(101)Acceleration due to gravity

[WPCareyDevil]

[SOLVED] (101)Acceleration due to gravity

Ok, this should be really strait forward. Its only problem C that I am not getting correct.

Homework Statement

An object is observed to fall from a bridge, striking the water below 3.36 s later.
(a) With what speed did it strike the water?

(b) What was the average speed during the fall?

(c) How high is the bridge?

Homework Equations

a) 9.8*3.36 = 32.928 m/s - correct

b) Vi+Vf/2 = 16.464 m/s - correct

c) ?

The Attempt at a Solution

Ok, acc due to gravity = 9.80m/s^2

So, in the first second, the object fell 9.8m. In the second second, it fell 19.6m. t=3 it fell 29.4m. t=3.36 it fell 32.928m.

Adding all of these would give me the total distance fallen (height of the bridge). The answer 91.728m is incorrect.

Am I attacking this the wrong way?

Thank you!

 

[Biest]

There is a formula for calculating the distance traveled due to acceleration... might want to give that a try

 

[||spoon||]

why not use x=ut+1/2at^2, where x is distance, u is initial velocity, a is acceleration and t is time.

 

[Battlecruiser]

Use the formula d = v_0 t + (1/2) a t^2

You know that initial velocity is 0. You know the time and you know the acceleration.

 

[JaWiB]

>So, in the first second, the object fell 9.8m. In the second second, it fell 19.6m. t=3 it fell 29.4m. t=3.36 it fell 32.928m.

If the object travels for 1 second at 9.8m/s, then it will travel 9.8m. Unfortunately, the speed in this case is not constant but increasing.

 

[WPCareyDevil]

I solved the problem with the above formula.

I was saying that it fell at 9.8m/s per second, thus it fell 9.8m in the first second, and then 19.6 meters in the second second, meaning it had fallen a total of 28.6m at the end of 2 seconds. I don't understand why this approach doesn't work, but all is well with the d=Vot+ (1/2)at^2 formula.

 

[Battlecruiser]

I was saying that it fell at 9.8m/s per second, thus it fell 9.8m in the first second, and then 19.6 meters in the second second, meaning it had fallen a total of 28.6m at the end of 2 seconds. I don't understand why this approach doesn't work, but all is well with the d=Vot+ (1/2)at^2 formula.

I think your approach ignores the fact that velocity is changing (Because there is acceleration) and it may be ignoring any initial velocity the object has already attained in the previous seconds.

 

[JaWiB]

No, it fell less than 9.8m in the first second because it is only traveling at 9.8m/s for an instant. It would only fall 9.8m in one second if it were actually traveling at 9.8m/s for the whole second, but it (presumably) starts with zero velocity. You actually have to take the area under the velocity-time graph, so for the first second it travels .5*(9.8-0) m or 4.5m (and for constant acceleration, this turns out to be average velocity multiplied by time)

 

[WPCareyDevil]

That makes perfect sense. Thank you all for the help!