(Should-be-easy) Physics Problem

[Tetra]

I'm not sure what i'm doing wrong for the following question:

 

"An object is dropped from a tower. If it had been thrown down at 60m/s, it would half taken have the time. What is the height of the tower? (Note that acceleration is 10m/s^2 here)"

 

So the parameters should be:

 

V (initial): 60m/s

V (final): 0m/s

A: 10 m/s^2

T: 6 s

 

And if I used the equation D = V (initial)*T + 1/2 a(t)^2

 

and subbed it all in, I ended up with 540m for the height. But the answer is supposed to be 320. Does anyone know what I did wrong?

[Cap'n Refsmmat]

Why is the final velocity 0 m/s? When it hits the ground it'll be traveling at some large velocity downward. If you want to take into account the fact that it'll stop when it hits the ground, you'll have to calculate the acceleration it experiences when it hits, which depends on how soft the grass is, how rigid the object is, and so on.

 

Best not to make the final velocity 0.

[Tetra]

So what could the final velocity be, then? And it's not necessary to calculate for the answer, anyways. You only need V (initial) for that equation. Are any of the other parameters (possible somehow) wrong?

[Cap'n Refsmmat]

I would leave out the final velocity altogether, as you say it's unnecessary.

 

How did you arrive at the 6 s figure for time?

[Tetra]

Well (and I think THIS is where I did it wrong)

 

A: 10 m/s^2

V (initial) = 60 m/s

 

I guess here I must've assumed that V (final) was 0, because then:

 

a = (V1 - V2) / T

10 = (60 -0) / T

T = g0 / 10

T = 6

 

Hey, should I just, I dunno, PM you or something just to make it easier? Than talking through a thread, that is...

[Cap'n Refsmmat]

Nah, that takes just as much effort as a thread, if not more.

 

Here's what I'd do. Set up your equations for the distance fallen by the object. You'd have [imath]y=y_0 t + \frac{1}{2}at^2[/imath], where [imath]y_0[/imath] is the height of the tower. y would be 0, because the object ends up on the ground.

 

Since y is 0 in both cases, you can set the two equations equal. One equation will use t, the other (for the object thrown down) can use [imath]\frac{1}{2}t[/imath]. Then solve.

[Tetra]

Oh hey, thanks! I set up the equations differently (had y be the height of the tower, and y0 be the velocity, since they're different for both), and got the right answer! Thanks man.

 

And if you wouldn't mind...could you help me with this other one too? Don't worry, I won't ask for any more

 

"From a point 70 m above the ground, an object is projected vertically upwards with a velocity of 25 m/s. How long will it take to reach the ground, assuming acceleration is 10 m/s^2."

 

Similar to the first question, except we have a distance... and we don't know when the object starts to fall.

[Cap'n Refsmmat]

That's not that bad. Set up [imath]y = y_0 t + \frac{1}{2}at^2[/imath] with y being 0, [imath]y_0[/imath] being the initial height, and a being gravity. (Make sure you make it negative.)