Answer:
v = -1.8t+36
20 seconds
360 m
40 seconds
36 m/s
The object speed will increase when it is coming down from its highest height.
Explanation:
Differentiating with respect to time we get
a) Velocity of the object after t seconds is v = -1.8t+36
At the highest point v will be 0
b) The object will reach the highest point after 20 seconds
c) Highest point the object will reach is 360 m
d) Time taken to strike the ground would be 20+20 = 40 seconds
![[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s](https://tex.z-dn.net/?f=%5Btex%5Dv%3Du%2Bat%5C%5C%5CRightarrow%20v%3D0%2B0.9%5Ctimes%202%5Ctimes%2020%5C%5C%5CRightarrow%20v%3D36%5C%20m%2Fs)
Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of 
e) The velocity with which the object strikes the ground will be 36 m/s
f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.
I think by using data collected by Tycho Brahe
4. The Coyote has an initial position vector of 
.
4a. The Coyote has an initial velocity vector of 
. His position at time 
is given by the vector
where 
is the Coyote's acceleration vector at time . He experiences acceleration only in the downward direction because of gravity, and in particular 
where 
. Splitting up the position vector into components, we have 
with
The Coyote hits the ground when 
:
4b. Here we evaluate 
at the time found in (4a).
5. The shell has initial position vector 
, and we're told that after some time the bullet (now separated from the shell) has a position of 
.
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where 
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for :
C.) i believe because the change in shape is changing the physical appearance
I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.
If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.
For the wave you described . . .
Frequency = (speed) / (wavelength)
= (6 m/s) / (12 m)
= 0.5 / sec
= 0.5 Hz .
(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)
