A stone is thrown vertically in the air..........???

Question

at a velocity of 16 m/s. One second later a 2nd stone is thrown up w/ the same velocity. At what time are the two stones at the same height?

I tried using y(t) = Vo*t+1/2 * a * t^2

set heights equal, therefore...

Vo* t + 1/2 * a * t^2= Vo* (t+1) + 1/2 * a * (t+1)^2

but got a negative answer, if someone could explain the right way to do this .....great.

ok i did that, but im getting the same wrong answer. just positive now. im getting 1.13. the answer should be 2.13.

and all the roots get cancelled, im only left w/ 9.8t=11.1

Answer 1

so far u r correct........now u just forgot 2 add 1 sec.....rememba u have to find the total time......1.13 is the time required for 2nd stone to travel.......so just add 1 to get d ans.

Answer 2

Remember that the way you have it set up the acceleration of gravity is a negative number. I am not sure you took this into account, but this could be the reason for your negative result. Also you should have two answers as this will lead to a quadratic, try the other root and see if you get a positive result.

Answer 3

what a bout this, try this.

try to imagine graph plotted in velocity-time.

for the 1st stone, v. (v with 1 dot) is 16m/s when the time is 0.
hence, (16,0)
gradient = -9.8m/s^2

you will get, (V.-16)= -9.8 (t-0)
simplify it you'll get an equation for v.

v.= -9.8t+16

for second stone, use v.. (v with two dots)
it has velocity 16 when time = 1s, hence
gradient = -9.8, (16,1)

then, you'll get v..= -9.8t+25.8

integrate v you'll get displacement. Hence,
integrate v. = s. = -4.9t^2 + 16t + c
when t = 0, s=0
hence, s.= -4.9t^2 + 16t

also, integrating v.. , you'll get s.. = -4.9t^2 +25.8t +c
when t = 1, s = 0
hence, c = -20.9
s.. = -4.9t^2 +25.8t - 20.9

they meet when s. = s..
-4.9t^2 + 16t = -4.9t^2 +25.8t - 20.9
hence, t = 2.13.... second

hope you'll appreciate.

Answer 4

They can't be together at the same height at any time.