How High Was the Building If a Rock Hits the Ground 43m Away?

[sirfederation]

1. A rock is thrown from the roof of a building, with an initial velocity of 10 m/s at an angle of 30 degrees above the horizonal. The rock is observed to strike the ground 43 m from the base of the building. What is the height of the building?


This is what I did:
First I set my origin at the top of the building so the height would be in the negative direction
Vxi [/size] = 10 cos 30 = 8.66 m/s
Vyi[/size] = 10 sin 30 =5 m/s

Xxf[/size] - Xxi [/size] = 1/2( Vi[/size]-Vf[/size])T
43-0 = 1/2 (8.66 + 8.66)T
43 = 8.66T
4.97 s = T

Vyf[/size] = 5 m/s -(9.8 m/s^2)(4.97s)
Vyf[/size] = -43.61 m/s

Yf[/size]-Yi[/size] = 1/2(Vi[/size] + Vf[/size])T
Yf[/size] = 1/2(5 -43.61)4.96
Yf[/size] = -95.75 m

What I was wondering is if I found Time which is T and the height correctly?

 

[gnome]

I'm having a hard time reading what you wrote there, but maybe this will give you what you need:

y_f - y_i = v_yi*t + (1/2)at²

Just remember that your v_yi is positive and a is negative.

 

[lavalamp]

Well I split everything up into horizontal and vertical componants, so first I will deal with the horizontal:

u = 10 Cos30 ms^-1
v = v
a = 0
s = 43 m
t = t

find the time taken (t) for the rock to travel 43m horizontally:

s = ut + 1/2 a t^2

43 = 10 t Cos30

43 = 5 ([squ]3) t

t  = 8.6 ([squ]3)

t  = 4.97 s


And now the vertical:

Then find out how far something can fall due to gravity in 4.97 s

u = 10 Sin60 = 5 ms^-1
v = v
a = 9.81 ms^-2
s = s
t = 4.97 s

s = ut + 0.5at^2

s = 5 * 4.97 + 0.5 * 9.81 * 4.97^2

s = 146 m

I'm not saying that what I've done is 100% correct, I could have made a silly error or something. And I'm not saying that you should do it this way, if you've been taught another way, do it like that if you feel more comfortable with it.
We get the same answer for the time of the drop, so it looks as though you made an error in one of the calculations after that but I haven't checked your working so I don't know.

 

[sirfederation]

Wouldn't the acceleration be -9.8 m/s^2 in lavalamps reply which would come out to -95.77

 

[lavalamp]

I took the downward direction as positive (maybe I should have mentioned that). That way, the value attained for s is actually the distance from the top of the building to the bottom of the building.

 

[HallsofIvy]

Xxf - Xxi = 1/2( Vi-Vf)T
43-0 = 1/2 (8.66 + 8.66)T
43 = 8.66T
4.97 s = T

This is incorrect. You are assuming that Vf= -Vi. If you were looking at vertical motion, and that you were looking at the same height both times, yes, the final velocity (down) would be the same as the initial (up) but in the opposite direction. Here, the speed horizontaly is a constant because there is no horizontal force.
You want Xxf- Xxi= Vi(T).

Using that value with what you did after that should give you the right answer.+

 

[sirfederation]

so if xxf[/size]-xxi[/size]=(Vi[/size])T
43m-0 = 10(m/s)T
4.3s = T

Then I plug it into the equation: yyf[/size]-yyi[/size]= Vyi[/size](T)+(1/2)(Ay[/size])(T)^2

yyf[/size]= 5(4.3)+(1/2)(-9.8)(4.3)^2
yyf[/size] =21.5-90.6
yyf[/size]=-69.1 m

I set my origin at the top of the building. Is this what it would come out to?

 

[lavalamp]

If you set your origin to the top of the building (like I did) then it would make sense to take the downward direction as positive (like I did). I would have thought that getting a negative answer would a dead give-away as a wrong answer.

 

[sirfederation]

I am posting the correct answer which is indeed -96 meters or 96 meters (determined by where you set your origin).