Physics help please PHYSICS please?

Question

Physics homework help please physics help?
1) Estimate the following.
(a) how long it took King Kong to fall straight down from the top of a 280 m high building
_____ s
(b) his velocity just before "landing"
_____m/s upward or downward? 

2) A helicopter is ascending vertically with a speed of 2.50 m/s. At a height of 110 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [Hint: The package's initial speed equals the helicopter's.]
_______ s

3) A baseball pitcher throws a baseball with a speed of 42 m/s. In throwing the baseball, the pitcher accelerates the ball through a displacement of about d = 3.4 m, from behind the body to the point where it is released. Estimate the average acceleration of the ball during the throwing motion.
________ m/s2


4) A stone is thrown vertically upward with a speed of 13.0 m/s from the edge of a cliff 65.0 m high. 
(a) How much later does it reach the bottom of the cliff? 
_______ s 
(b) What is its speed just before hitting? 
_______ m/s 
(c) What total distance did it travel? 
______ m

royal_knight2005 · Accepted Answer

1)-
s=at^2/2+v1t
v1=0
s=at^2/2
2s=at^2
t^2=2s/a
a=9.81 m/s^2
2s=2*280=560 m
t^2=560/9.81=57.0846
t=7.555<----- time needed to fall down from 280 m high

Velocity =V
V=at+v1<----- V1=0
V=at= 9.81*7.555=74.11882 m/s<---- speed at the moment of impact 


2)-
because the speed of the helicopter is opposite to the falling speed we do the following
s=at^2/2 +v1t

v1=-2.5 m/s
s=110 m
a=9.81 m/s^2
110=9.81*0.5*t^2+(-2.5*t)
110=4.905t^2 - 2.5t
0=4.905t^2 - 2.5t -110 <--------- use your calculator   
t=--4.4.... s <-------- to be neglected cause time cant be negative 
t= 4.997313955 s


3)-
s=(v+v1)t/2
t=2s/(v+v1)
t= 3.4*2/42+0=6.8/42=0.1619 s
a=(v-v1)/t
a=42/0.1619=259.4118 m/s^2



4)-
we need to work out the needed time to come to zero velocity upwards
t= (V - V1)/a
V= 0
V1=13 m/s
a= -9.81 m/s^2
t= -13/-9.81=1.32517 s<---------- time @1
now we need to know the distance upwards
s=(V+V1)*t/2
s=(13*1.32517)/2= 8.6136 m

now add that to the cliff height to get the complete distance that to be traveled by the stone downwards
s=8.6136 + 65 
s= 73.6136 m

the time to travel that distance is 
s=at^2/2+V1t<------- V1 =0 
73.6136 =9.81*0.5*t^2
t^2=15.008 s^2 
t=3.874 s<-------------- time @2

so total time = time@1 +time@2= 5.19917912 s 




maybe!!!

Anonymous · Answer

I'll do the first for you... you can try the others yourself, otherwise you'll never learn.

1) a) S = +0.5 at²

t = 2S /a

t = 560/9.8

t = 57 seconds.

b) v = u + at
v = 0 + 9.8 x 57

v = 558.6 m/s (downwards, obviously)

Physics help please PHYSICS please?

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