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Ball thrown from building

  • Thread starter Eiano
  • Start date
12
0
Hello!

I've been working on this problem for a while and our book doesn't have a solutions manual. It has 4 parts to it and I've solved 2 already, (I think) and the last 2 are just getting to me.

-A ball is thrown from the top of a building and is given an initial velocity of 30 m.s upward. The building is 60M high.

1). Determine the time needed for the stone to reach its max height
vf=at+vi
-30=-10t, t=3

2). Maximum height?
d= .5AT^s +viT
.5(-10)(3)^2+30(3)
= -5(9)+ 90
=15M

Velocity of the ball just before it hits the ground and the total time the ball is in the air?

What i did was
x=voT +.5AT^2 and solved for T but that cant be right because the answer i get is less that 7 seconds, and question following this says find the velocity and position at 7 sec. so it has to be AT LEAST 7 sec.

then i substituted that t for v=vo+at. so really, the 4th question is not the problem it's the 3rd.

Any help is appreciated
 
Eiano said:
Velocity of the ball just before it hits the ground and the total time the ball is in the air?

What i did was
x=voT +.5AT^2 and solved for T but that cant be right because the answer i get is less that 7 seconds, and question following this says find the velocity and position at 7 sec. so it has to be AT LEAST 7 sec.

then i substituted that t for v=vo+at. so really, the 4th question is not the problem it's the 3rd.
Why do you have voT? vo is supposed to be zero (this is the velocity in the beginning).
So you need only x = ½gt^2 and after that v = at.

(Or you can use conservation of mechanical energy mgh = ½mv^2).
 
12
0
KingOfTwilight said:
Why do you have voT? vo is supposed to be zero.
So you need only x = ½gt^2.

isnt Vo the initial velocity which is 30 m/s?
 
Eiano said:
isnt Vo the initial velocity which is 30 m/s?
I shouldn't be doing this in the night... I'm making mistakes all the time, sorry.
You know the time to go to the max height. Then you can calculate the time to drop from that height to the ground and use the knowledge in my 1st post.

EDIT:

After clearing my mind: yes you can use the one below too.
Check the answer by doing the both approaches.
 
Last edited:
12
0
KingOfTwilight said:
Oh, sorry, yes it is. So, you will have to use x = x0 + v0T + ½aT^2.
x0 is the height of the building.

ahh i forgot the x0, you are right, thank you!!!!
 
12
0
KingOfTwilight said:
I shouldn't be doing this in the night... I'm making mistakes all the time, sorry.
You know the time to go to the max height. Then you can calculate the time to drop from that height to the ground and use the knowledge in my 1st post.

EDIT:

After clearing my mind: yes you can use the one below too.
Check the answer by doing the both approaches.

it's okay! at least there was someone out there to help me. again, i appreciate it!!!
Thanks. :)
 

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