Calculate when displacement are equal

[ahmedb]

Homework Statement

One stone is dropped from the the top of a tall cliff, and a second stone with the same mass is thrown vertically from the same cliff with a velocity of 10.0 m/s [down], 0.50seconds after the first. Calculate the distance below the top of the cliff at which the second stone overtakes the first?

Homework Equations

v=0
d= -1/2(a)(t)^2

v=10
d=10(t)-1/2(a)(t)^2

The Attempt at a Solution

we have two unknowns, how can we solve it??

 

[SammyS]

Homework Statement

One stone is dropped from the the top of a tall cliff, and a second stone with the same mass is thrown vertically from the same cliff with a velocity of 10.0 m/s [down], 0.50seconds after the first. Calculate the distance below the top of the cliff at which the second stone overtakes the first?

Homework Equations

v=0
d= -1/2(a)(t)^2

v=10
d=10(t)-1/2(a)(t)^2

The time in this equation should be (t - 0.50), if t is the time in the previous equation .

The Attempt at a Solution

we have two unknowns, how can we solve it??

Two equations in two unknowns.

 

[ahmedb]

i still don't get the answer, can you explain the steps please

 

[SammyS]

i still don't get the answer, can you explain the steps please

First rewrite the second equation,

d=10(t)-1/2(a)(t)² ,

using (t - 0.5) in place of t. (This is because the second stone is in the air for 1/2 second less time than the first stone.)

Then "FOIL" the (t - 0.5)², multiply out everything in the expression and then collect terms.

See what you have then.

 

[ahmedb]

First rewrite the second equation,

d=10(t)-1/2(a)(t)² ,

using (t - 0.5) in place of t. (This is because the second stone is in the air for 1/2 second less time than the first stone.)

Then "FOIL" the (t - 0.5)², multiply out everything in the expression and then collect terms.

See what you have then.

so after you foil, you make the displacement equal to each other, and then solve for t, and then plug "T" into the original equation?

 

[SammyS]

so after you foil, you make the displacement equal to each other, and then solve for t, and then plug "t" into the original equation?

Yes.

 

[ahmedb]

Yes.

can you also do: d= -1/2(a)(t+.5)^2

cas the ffirst stone is .5 seconds in air longer than the second stone

 

[SammyS]

can you also do: d= -1/2(a)(t+.5)^2

cas the ffirst stone is .5 seconds in air longer than the second stone

You can do either

t for the first stone and (t - 0.5) for the second stone,

OR

(t - 0.5) for the first stone and t for the second stone,

but not both (t + 0.5) and (t - 0.5).

 

[ahmedb]

You can do either

t for the first stone and (t - 0.5) for the second stone,

OR

(t - 0.5) for the first stone and t for the second stone,

but not both (t + 0.5) and (t - 0.5).

don't you mean "t for the first stone and (t "+" 0.5) for the second stone,"?

 

[SammyS]

don't you mean "t for the first stone and (t "+" 0.5) for the second stone,"?

No. No matter how you express the time for each stone the time for the second stone is always 0.5 s less than the time for the first stone.

Therefore, the time for the first stone is always 0.5 s more than the time for the second stone.

 

[ahmedb]

No. No matter how you express the time for each stone the time for the second stone is always 0.5 s less than the time for the first stone.

Therefore, the time for the first stone is always 0.5 s more than the time for the second stone.

(t - 0.5) for the first stone and t for the second stone,

for that cas the time for the first one is 5 less than the time for the second stone.

so is this correct "(t + 0.5) for the first stone and t for the second stone,"?

 

[SammyS]

(t - 0.5) for the first stone and t for the second stone,

for that cas the time for the first one is 5 less than the time for the second stone.

so is this correct "(t + 0.5) for the first stone and t for the second stone,"?

Then, while the first stone falls for say 1 second, the the second stone would fall for 1 + 0.5 seconds. That's just not correct.

 

[ahmedb]

ohh lol, now i get it, thanks :D