two stonees are dropped from the top of a tower at half a second apart the time after dropping the first stone at which the distance between the two stones is 20m is??

Question

experimentx · Answer

?? i did not understand your question ??? is it asking time or distance ??? also is distance between two stones 20m ??

anonymous · Answer

its about time??

anonymous · Answer

options are (a)  4s   b 3.75s         c   4.25s      d   2s  the answer is c can u solve it??

experimentx · Answer

$ s_1 - s_2 = 20 = (u_1t + 1/2 * at^2) - (u_2(t-1/2) + 1/2 * a(t-1/2)^2) $, u1, u2 = 0, a = g

experimentx · Answer

http://www.wolframalpha.com/input/?i=solve+t^2+-+%28t-1%2F2%29^2+%3D+4

anonymous · Answer

i did nt understand this +1/2∗a(t−1/2)2 can u clear it

anonymous · Answer

take g=10m/s2

experimentx · Answer

the time factor is decreased by half for second stone, i.e. T = t - 1/2

amistre64 · Answer

how far does the first stone drop in half a second?

anonymous · Answer

how it came  t^2 - (t-1/2)^2 = 4???

experimentx · Answer

well ... i simplified it greatly!!! .... use above relation!!!

amistre64 · Answer

after half a second, the first stone can be thought of to have an initial velocity compared to that of the second stone. .....

p(t) = 9.8 t^2
v(t) = 4.9 t   ; at t=1/2 = 2.45

p1(t) = 9.8 t^2 + 2.45t
p2(t) = 9.8 t^2

when p1(t) - p2(t) = 20  we should be good

amistre64 · Answer

plus half a second :)

experimentx · Answer

see you later ... !!!

amistre64 · Answer

its best to interact in the post and not the messages of you have questions

amistre64 · Answer

the position of the 1st stone under gravity can be modeled by the equation

hmm, my gravity was off a little, but concept should be sound

anonymous · Answer

take g=10m/s2

amistre64 · Answer

the equation for position is:$$p(t)=\frac{1}{2}g\ t^2+V_ot+H_o$$

the derivative of position is velocity; $$v(t)=gt+V_o$$

amistre64 · Answer

when the time = .5; the velocity of the first stone is at:

10+.5 = 5  right?  so we adjust the equation and compare it to the second stone

amistre64 · Answer

10*.5 = 5  cant type to save my life tonight

amistre64 · Answer

$$p_1(t)=\frac{1}{2}g\ t^2+V_{o}t$$
$$p_2(t)=\frac{1}{2}g\ t^2$$

when the position between 1 and 2 = 20, we can solve for t

$$p_1(t)-p_2(t)=20$$
$$\frac{1}{2}g\ t^2+V_{o}t-\frac{1}{2}g\ t^2=20$$
$$V_ot-20=0$$
$$t=\frac{20}{V_o}$$
total time elpsed is half a second more then

amistre64 · Answer

20/5 = 4 + .5 = 4.5

but thats if you use g=10 instead of 9.8