Kinematics.?

Question

Please help me with this one, we had a test and no one (!) could figure that problem out. (9th grade physics)
btw I don't live in America so I had to translate it..sorry

object A is released from a top of a building and falls. At the same moment object B is fired from the ground upwards, with a starting velocity that allowes it's apex to be the building height.
When the two bodies pass by, B's velocity is 12.25 m/s .

a. What is A's velocity at the moment they pass by?

b. What is B's starting velocity?

c. How tall is the building?

d. When the objects pass by, what is their height?

And again, sorry if I translated it wrong or something..

Answer 1

Ok, the only known data you have given is B's velocity when both objects pass by.

Assuming the B initial velocity is VB1, then final VB velocity is 0 and that is when B reaches to apex of the building height.
VB final = VB initial –g*t, then VB1 = g*t = 9.8t
When B's velocity is = 12.25, then A and B meet in the air
The Height of the building is = Y = VB1*t –0.5*9.8*t^2
Y = 9.8t^2 -0.5*9.8*t^2 = 4.9t^2
Y also = to VA1*t +0.5gt^2. VA1 = 0, then: Y =4.9t^2
VA final = VB1 initial; VA final = g*t = 9.8t
The time it takes B to meet with A is the same and let that time be equal to t x
V B x = -g*t x = VA x = g*t x; 12.25 = 9.8t x
t x = 12.25/9.8 = 1.25 second
Therefore Both A and B pass each other at time = 1.25 s and that is when both VB x = VA x = 12.25 m/s
Now taken B at the time of passing by A, the equation formulated from the data extracted is:
V final = VB x -9.8t2, where t2 is the time starting from the time when both A and B meet until B reaches to the apex of the building, or until A reaches the ground.
12.25 = 9.8t2; then t2 = 1.25 seconds
Total time for B to reach to the apex of the building is 1.25 +1.25 = 2.5 seconds which is the total time for A and B to reach the ground for A and the apex of the building for B
Since total time extracted from the equations of motion = 2.5 seconds then Y = 4.5*2.5^2 = 28.125 meters
VA final = VB initial = 9.8*2.5 = 24.5 m/s


The four unknowns you requested are:

VA velocity at the time it passes B is = to VB velocity
=12.25 m/s

VA final = VB initial = 9.8*2.5 = 24.5 m/s

The building is 28.125 m tall

The height when both A and B passes by = 4.9 * t x^2 =4.9*1.25^2= 7.656 meter from the top of the building . Or 28.125 -7.656 =20.6 m from the ground.

I hope I assisted in finding the solution of this wonderful problem!

Answer 2

I can tell just from the problem that you do not live in America. The solution lies in recognizing that B only had the velocity to reach the top of the building.

a.) As B goes up to the top at every point its velocity is equal in magnitude to the velocity of A were it falling at that point. This is because B's apex would be the top of the building. Imagine that B made it to the top, after that point B would act exactly like A, so it must have acted the exact opposite of A while climing upward. Thus A's velocity at that point must have been 12.25 m/s since they ARE at the same point.

b.) If A had a velocity of 12.25 m/s moving from rest then B must have had a velocity of V-12.25 m/s since gravity is the only force acting on these objects. Therefore B started at a velocity of 24.5 m/s.

c.) The height of the building is the sum of 2 distance formulas. First find the time it took to accelerate A to 12.25 m/s which is (12.25m/s)/(9.80m/s/s) = 1.25 seconds. Then sum 1/2*(9.80m/s/s)(1.25s)^2 + 1/2(24.25m/s+12.25m/s)*(1.25s) which = 30.63 m.

d.) They pass by at the height where they meet. A trivial calculation. 1/2(1.25s)(24.5+12.25) = 22.99 m above earth.

Answer 3

We'll put the origin at the bottom of the building and call "up" the positive y direction. The height of the building is H. The acceleration of gravity, g, is 9.8 m/s² down.

Start with the motion of object A. Its position and velocity are given as

ya = H - 1/2 g t²
va = -gt

Now for the motion of B.

yb = vob t - 1/2 g t²
vb = vob - gt

We know that vob is sufficient to get B just to the top of the building...

2 g H = vob²
H = vob² / 2g

Now look at the instant at which they pass, time t. We know that at this instant, yb = 12.25 m/s. Alas, we do NOT know if B is rising or falling (that is, do we have 12.25 m/s or -12.25 m/s?). So we have to consider them both...

+ or - 12.25 m/s = vob - gt

At this time the two objects have the same height, so ya = yb at that time:

ya = yb
H - 1/2 g t² = vob t - 1/2 g t²
H = vob t

Combine this with the previous result of H = vob² / 2g...

vob t = vob² / 2g
t = vob / 2g

Combine THAT with what we know about the speed of B...

+ or - 12.25 m/s = vob - gt

Plug in vob / 2g for t and solve for the two possible values of vob. That should be more than enough to get you started...