physics freefall question?

Question

A girl throws a ball from a window to her sister in a window 4 meters above. Her sister catches it 1.5 seconds later. What was the initial velocity and what was the velocity just before the ball was caught?

Answer 1

A tricky question. Apparently, the ball was tossed upward; physics predicts that regardless of the horizontal velocity, the ball would reach a certain verticle level at the same time as if it were simply dropped. So you need to set up a problem like this: Let th be time at which the ball reaches its highest elevation. Then to reach zero velocity,
0 = vo - a*th. [a= 9.8 m/sec^2]
From this point, we have free fall. So
vf = a * (1.5-th)
These can be combined to express vo and vf in terms of th.
We can obtain more info from positional eqtns:
On the way up xup = vo*th - (a/2)*th^2
On the way down, xup+4= -(a/2)*(1.5-th)^2
You can eliminate xup from both equations to get vo in terms of th.
Then you can solve. Go for it/

Answer 2

Vf = at + Vi

Vf = final velocity
a = acceleration
t = time
Vi = intial velocity


Vf = (-9.8)(1.5) + Vi
Vf = -14.7 + Vi


Vf^2 = 2ad + Vi^2

d= displacement

(-14.7 + Vi)^2 = 2(-9.8)(4) + Vi^2
216.09 - 29.4Vi + Vi^2 = -78.4 + Vi^2
216.09 - 29.4Vi = -78.4
-29.4Vi = -294.49
Vi = 10.02 m/s

so the initial velocity of the ball is 10.02 m/s

to find the velocity when the ball is catched, plug 10.02 for Vi
Vf = -14.7 + Vi
Vf = -14.7 + 10.02
Vf = -4.68m/s

Answer 3

Let u and v be the initial and final velocity.
Since a = [v-u] /t

- 9.8 = [v-u] / 1.5.

[v-u] = - 14.7m/s

[v+u]t / 2 = s {distance = average velocity x time}

[v+u] = 4 x 2 /1.5 = 5.333m/s

[v+u] - [v-u] = 2u = 20 .0333

u = 10.017 m/s up

v = - 4.68m/s down

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Or using

s = ut - 1/2 gt^2

4 = 1.5 u - 0.5 x 9.8 x 1.5^2

u = 10.017 m/s and

v = u - gt

v = 10.017 - 9.8 x 1.5 = - 4.683 m/s