# Freefall body problem

1. Aug 19, 2015

### VaioZ

1. The problem statement, all variables and given/known data
You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at a point even with the roof railing with an upward speed of 15.0 m/s; the ball is then in free fall. On its way back down, it just misses the railing.
Find (a) the position and velocity of the ball 1.0 s and 4.0 s after leaving your hand;
(b) the velocity when the ball is 5.0 m above the railing;
(b.1) How long will the ball reach 5 m
(c) the maximum height reached and the time at which it is reached; and
(d) the acceleration of the ball when it is at its maximum height.

2. Relevant equations
Freefall equation like Vf=Vo-gt and etc

3. The attempt at a solution
T at 1
a.) ΔY = (15 m/s)(1 s) - (1/2)(9.8 m/s^2)(1 s)^2
ΔY = 10.1 m

T at 4
ΔY = -18.4 m

V at 1
Vf = (15 m/s) - (9.8 m/s^2)(1)
= 5.2 m/s

V at 4
Vf = -24.2 m/s

b.) Vf^2 = Vi^2 - 2gΔy
Vf^2 = (15 m/s)^2 - (2)(9.8 m/s^2)(5 m)
Vf^2 = 11.27 m/s

b.1) (11.27 m/s) = (15 m/s) - (9.8^2)(t)
t = 0.38

5 m = (15 m/s)(t) - (1/2)(9.8 m/s^2)(t)
t = 2.68

I'm a bit confuse here I got 2 different answer in this part. Really need a big help here

c.) a = Δv/Δt -> t = Δv/a -> t = (15 m/s)/(9.8 m/s^2)
tmax = 1.5306

Δymax = (15 m/s)(1.5306 s) - (1/2)(9.8 m/s^2)(1.5306 s)^2
y = 11.48

d.) The answer is -9.8 m/s^2 right? Constant?

2. Aug 19, 2015

### Sobhan

For question b.1) you wrote 5=15t-9,8*0.5t while it should be 5=15t-9.8*0.5T^2

3. Aug 19, 2015

### VaioZ

lol im such a stupid one

is my c.) and d.) right?

4. Aug 19, 2015

### haruspex

Yes, c and d are right.