# Need help solving

#### Rkaz

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

#### studiot

What data (distances, velocities, accelerations etc) do you know immediately from the question ?

#### Rkaz

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

#### studiot

There was a whole lot of data given in 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?

#### Woody

Is one object dropped from the top of the building,
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 s ?

As Studiot points out, be careful with your sign convention
If the acceleration (g) is positive down, then all your other values must also be positive down (and up is negative).
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 chose the bottom of the building to be at 0m, then (using up as negative) the top of the building is at -30meters
or if you chose the top of the building to be at 0 meters, then the bottom of the building is at +30 meters

You could use down as negative and up as positive, (but then g = -9.81m/s^2 and the initial velocity is +10m/s)
it is a matter of personal preference, and will give the same answer (but with the opposite sign)

#### studiot

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

#### Doniyor

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
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 $$\displaystyle g=9.81 m/s^2$$.
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 t : $$\displaystyle 30=10t-15+3*9.81/2-9.81*2.25/2$$ and $$\displaystyle 30=-15+24.715*t-11.03625$$ and finally, at $$\displaystyle t\approx 2.27 s$$ they meet. Substitute this 2.27 s either to $$\displaystyle y_1$$ or $$\displaystyle y_2$$ to get that hight they meet. $$\displaystyle y_1=30-9.81*(2.27)^2/2=4.72$$ meters. The answer is 4.72 meters.