Question

a stone is dropped from a captive balloon at an elevation of 1000ft,. two sec later another stone is projected vertically upward from the ground with a velocity of 248ft per sec. if g is 32ft per sec squared when and where will the stones pass each other?

Answer 1

wow is that a physics question

Answer 2

obviously..yes :)

Answer 3

this is the first time i ever saw a physics question lol cuz i study business stream

Answer 4

can u write a equation for the first stone distance travelled by it in t seconds

Answer 5

i know that:
\[V _{01}=0 ft/s\]
\[V _{02}=248 ft/s\]
\[g=32.2 ft/s^2\]
\[t _{1}=t _{2}+2\]

Answer 6

@unkabogable can u?

Answer 7

So for the first stone distance travelled by it in t seconds is = ut + 1/2 at^2
= 16t^2

Answer 8

right?

Answer 9

yes.

Answer 10

yes..

Answer 11

now find, for the first stone distance travelled by it in 2 seconds

Answer 12

can u?

Answer 13

yeah...wait:
so i have these eq.
\[Y _{1}=1000-Y _{2}\]
\[Y _{2}=V _{02}t _{2}-1/2(32.2)t _{2}^2\]
then i'll substitute and i'll verify my answer later

Answer 14

Let me do it then,
for the first stone distance travelled by it in 2 seconds=16*2^2 = 64ft
now velocity of it after two seconds = (o^2 + 2*32 *64)^1/2 = 64ft/s

Answer 15

Now,Now,
after two seconds,
height of the stone after t seonds =1000-64-( ut+1/2 at^2)
= 936-64t - 16t^2

Answer 16

then for the next stone:
height of the stone after t seconds = ut + 1/2 at^2
= 248t + 1/2 (-32) t^2
=248t - 16t^2

Answer 17

Now, when the the stones pass each other ,
936-64t - 16t^2=248t - 16t^2
936=312t
t=3seconds

Answer 18

so the stone will pass each other at 3+2 seconds = 5 seconds

Answer 19

got it?

Answer 20

yes..thanks again!! :)

Answer 21

WElcome

Answer 22

No, can u find this
where will the stones pass each other

Answer 23

Now

Answer 24

(248)(5-2)\[(248)(5-2)- \frac{ (32.2(5-2)^2) }{ 2} = 600 ft\]
??

Answer 25

Yep........ 600 ft above the ground