Solving Projectile Motion Problems: Intersecting Balls at Different Heights

[Max0007]

Homework Statement

The top of a 10-storey building (42 m) , someone throws a ball (A) down with a
initial speed of 2.5 m / s. You are on the second floor (8 m window above the ground) and
You throw a ball (B) upward with an initial speed of 22 m / s.
a) If both balls are thrown together after how much time do they intersect ?
b) It is hoped that the two balls, launched with the same speed and in the same direction , strike
the soil simultaneously. Which bullets (A or B) is to be launched first? how
time before the other ball should it be launched ?

Homework Equations

Initial Speed for A: 2.5 m/s
height B initial: 42 m
height B final: 0 m
Initial Speed for B: 22 m/s
height B initial: 8 m
height B final: 0 m

The Attempt at a Solution

a)
height final= height initial + Initial speed *(final time - initial time) - g/2 *(final time - initial time)^2
0=8 + 22m/s * (time) - 9.81/2 * (time)^2
0=42 + 2.5m/s * (time) - 9.81/2 * (time)^2

To find the the time when it intersect I did this:
42 + 2.5m/s * (time) - 9.81/2 * (time)^2 = 8 + 22m/s * (time) - 9.81/2 * (time)^2

and found 1.74seconds

For b)

I found at what time they both hit the ground.

0=42 + 2.5m/s * (time) - 9.81/2 * (time)^2 // A hits the ground in 3.19 secs
0=8 + 22m/s * (time) - 9.81/2 * (time)^2 // B hits the ground in 4.82 secs

I want to trow B first because it takes longer than A to hit the ground.

Now since B hits the ground in 4.82 secs I will replace 4.82 sec in this 42 + 2.5m/s * (time final - time initial) - 9.81/2 * (time final - time initial)^2 , I will replace it in time final. So I will have to find the time initial. I want A to hit the ground the same time as B.

42 + 2.5m/s * (4.82 - time initial) - 9.81/2 * (4.82- time initial)^2=0

I find 2.14 secs.

Can you guys verify that its all good please ? I really appreciate it :)

 

[haruspex]

42 + 2.5m/s * (time) - 9.81/2 * (time)^2 = 8 + 22m/s * (time) - 9.81/2 * (time)^2

Which way was the ball thrown from the top? (Same problem in part b.)
There is an easier way to solve this. Consider a reference frame in free fall.

For b)
I found at what time they both hit the ground.
0=42 + 2.5m/s * (time) - 9.81/2 * (time)^2 // A hits the ground in 3.19 secs
0=8 + 22m/s * (time) - 9.81/2 * (time)^2 // B hits the ground in 4.82 secs

 

[Max0007]

Which way was the ball thrown from the top? (Same problem in part b.)

There is an easier way to solve this. Consider a reference frame in free fall.

"Which way was the ball thrown from the top? (Same problem in part b.)"
downward so its -2.5m/s

But for part b) I realized I did not do the right thing. I don't know how to actually do it ;(

 

[haruspex]

downward so its -2.5m/s

Right. The easier way, as I wrote, is to consider a reference frame in free fall. In that frame, each ball has a constant velocity. So how long do they take to meet?

But for part b) I realized I did not do the right thing.

Quite so.
You started out well, finding the time each takes. From there it was an extremely short step to the answer, but you wandered off course.

 

[Max0007]

Right. The easier way, as I wrote, is to consider a reference frame in free fall. In that frame, each ball has a constant velocity. So how long do they take to meet?

Quite so.
You started out well, finding the time each takes. From there it was an extremely short step to the answer, but you wandered off course.

Time B - Time A? only that :O

 

[haruspex]

Time B - Time A? only that :O

That'll do it.