Answer:
a.After
second Mr Comer's speed

b.Distance travelled by Mr.Comer in
seconds

Explanation:
a. Lets recall our first equation of motion 
Now we know that
,
and

Plugging the values we have.




Then Mr.Comer's speed after
sec

b.
Lets find the distance and recall our third equation of motion.

So
distance covered.
Dividing both sides with 2a we have.

Plugging the values.


So Mr.Comer will travel a distance of
.
Answer:

Explanation:
given,
velocity of particle 1 = 0.741 c to left
velocity of second particle = 0.543 c to right
relative velocity between the particle = ?
for the relative velocity calculation we have formula

u_x = 0.543 c
v_x = - 0.741 c




Relative velocity of the particle is 
Answer:
h=18.05 cm
Explanation:
Given that
m= 25 kg
K= 1300 N/m
x=26.4 cm
θ= 19.5 ∘
When the block just leave the spring then the speed of block = v m/s
From energy conservation



By putting the values


v=1.9 m/s
When block reach at the maximum height(h) position then the final speed of the block will be zero.
We know that

By putting the values

h=0.1805 m
h=18.05 cm
The angular speed is defined as:
<h2> ω=

</h2>
where


