How do I calculate the height of a tower using motion in one dimension?

[agm2010]

I have a few problems I'm having trouble with. If I can get some help with this one I should be able to figure out the rest I have.

1. A rock is thrown downward from the top of a tower with an initial speed of 12 m/s. If the rock hits the ground after 2.0 s, what is the height of the tower? (neglect air resistance).

Known:
Vi=12m/s
\Delta t=2 s
a=9.8 m/s

Relevant equations:
Vf^2=Vi^2+2a(\Delta y)
My attempt:
\Delta y=Vf^2-Vi^2-(2a)
\Delta y=0-144-(2*9.8 m/s^2)
\Delta y=-163.6
Height=163.6 m

I'm not sure if this is right. I think I run into trouble when I rearrange the formula. Multiple choice answer D is 63 m, so it's either a typo or I'm doing something wrong. A little guidance please? Thank you!

 

[rl.bhat]

If a = b + gc, then
c = (a - b)/g
So your Δy formula is wrong.

 

[agm2010]

Ah, ok. I see what I did.

Corrected formula:
\Delta y=Vf^2-Vi^2/2aI also need to first find the final velocity:

Vf=Vi+a(\Delta t)
Vf=12 m/s + 9.8 m/s^2 (2 s)
Vf=31.6 m/s

Now for \Delta y:

\Delta y = (31.6m/s^2) - (12 m/s^2) / 2(9.8m/s^2):
\Delta y = 43.6 m
Height: 44 m

So it turned out to be a silly mistake. At least I learned from it. Thanks rl.bhat!

 

[kNYsJakE]

You can also derive the formula that you need.

Here is an alternative way to solve this problem. This will help.

Given that the acceleration is 9.8 m/s^2 we know that

\ddot{x}=9.8 m/s^2

and given that the initial velocity is 12 m/s, we can get

\dot{x}=9.8t + 12

and finally, setting the initial point as 0 m

x=\frac{9.8}{2}t^2+12t

Now you can plug in t=2 sec, and you get

x=43.6 m

A simple calculus trick. This will help you solving more complicated problems later, without any formulae memorized. =)