physics problems. Urgent!!!

1. student drops a text bool from a 4th story window 13m above the ground. how fast will the book be traveling when it hits the ground?
By v^2 - u^2 = 2as,
Here u = 0m/s, s = 13m, a = g = 10m/s^2
v^2 = 2(10)(13) = 260
v = 16.12 m/s
By v = u+ at
16.12 = 10t
t = 1.61s
Therefore, the book will travel 1.61s to reach the ground at a speed of 16.12m/s
2. a baseball is thrown straight up from the surface of the earth with a speed of 45m/s. what will the displacement of the baseball be after 3s?
By s = ut - 1/2 gt^2
s = 45(3) - 1/2(10)(3)^2
s = 90m
Therefore, the baseball will be at a a height of 90m

1. By equation of motion ( v^2=u^2+2as)
v^2=0^2+2(10)(13) accelaration=10mS-2 (take downwards as positive)
v=16.1245155ms-1
Therefore the velocity when hitting the ground is 16.1ms-1.

2. accelaration= -10ms-1 (take upwards as positive)
By equation of motion ( s=ut+1/2at^2)
s=45^3+(-10)3^2/2
s=135-45
s=90m
Therefore the displacement will be 90m.

(By Y.Y)

2006-11-02 18:29:50 補充：
First, you should find out which information the question has given.An example in Q1.You are given s=13ma=10m ( take downwards as positive)u=0ms-1 And you are going to find velocity.So you should use the equation with s, a, u and v.

1. Vi=initial velocity= 0 m/s, d=displacement=13m, a=gravity force=acceleration= -9.8m/s^2~~~ find Vf=final velocity=?
Vf^2=Vi^2+2ad
Vf^2=2(-9.8 m/s^2)(13m)
Vf^2=2(-127.4 m^2/s^2)
Vf^2= -254.8 m^2/s^2
Vf= √-254.8 m^2/s^2
Vf= ±15.96 m/s

2. Vi=initial velocity= 0 m/s, a=gravity force=acceleration= -9.8m/s^2, t=time=3s, Vf=final velocity=45 m/s~~ find d=displacement=?

d=Vi×T+1/2at^2
d=0+1/2(-9.8 m/s^2)(3s)^2
d=1/2(-88.2m)
d=±44.1m

2006-11-02 16:38:55 補充：
question #2:the initial velocity should be 45 m/s, not 0 m/s....sorry about that~~displace the variable of Vi by 45 m/s...then u should find the answer

2006-11-02 16:39:47 補充：
answer for question #2 is 90.9 m

1. v=mh