- Thread starter Rkaz
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that was the whole question, nothing else given. Take gravity as 9.81m/s^2What data (distances, velocities, accelerations etc) do you know immediately from the question ?

Such as what was the initial velocity of the first object ?

What was the initial velocity of the second object?

If the first object falls 'm1' metres, how far does the second object rise at intersection?

If the first object falls for t1 seconds to intersection, how long is the second object rising to this point?

Since you are taking g as +9.81 m/s what are the implications for your sign convention?

and the other thrown up from the bottom of the building?

You should end up with two versions of the \(\displaystyle s=ut +1/2at^2\) equation

when do these two equations give the same value for

As Studiot points out, be careful with your sign convention

If the acceleration (g) is positive down, then

i.e the initial velocity of the thrown object is negative

Also be careful about how you define your zero point for distance:

If you

or if you

You could use down as negative and up as positive, (but then g =

it is a matter of personal preference, and will give the same answer (but with the opposite sign)

Yes when since the two distance will have the opposite sign?when do these two equations give the same value for?s

We can start with writing equations of coordinates of two objects \(\displaystyle y=y_0+v_0 t+a t^2/2\). For the first object \(\displaystyle y_1=30-g t^2/2\), and for the second object \(\displaystyle y_2=10(t-1.5)-g(t-1.5)^2/2\), where \(\displaystyle g=9.81 m/s^2\).An object is dropped from rest, from the top of a 30m high building. 1.5 seconds later, a second object is thrown upwards with an initial velocity of 10m/s at what height will the objects pass each other

Passing each other means that their coordinates are equal to each other, e.g. \(\displaystyle y_1=y_2\). That way we have an equation for unknown time moment when they pass each other: \(\displaystyle 30-9.81 t^2/2=10(t-1.5)-g(t-1.5)^2/2[\). Let's expand \(\displaystyle (t-1.5)^2=t^2-3t+2.25\) and get \(\displaystyle 30-9.81 t^2/2=10(t-1.5)-9.81(t^2-3t+2.25)/2\). We can cancel out \(\displaystyle 9.81 t^2/2\) from both sides, after the opening the brackets on the right-hand side. Then we get \(\displaystyle 30=10(t-1.5)-9.81(-3t+2.25)/2\). We can simplify this equation and solve it w.r.t. time