Answer:
1) The greatest height attained by the ball equals 20.387 meters.
2) The time it takes for the ball to reach 15 meters approximately equals 1 second.
Explanation:
The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.
thus using third equation of kinematics we obtain the height attained as

where
'v' is the final speed of the ball
'u' is the initial speed of the ball
'a' is the acceleration that the ball is under which in this case equals 9.81 
's' is the distance it covers
Thus for maximum height applying the values in the equation we get

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

Answer:

Explanation:
d = Diameter of spot = 30 μm
r = Radius of spot = 
P = Power of the laser = 
A = Area = 
Intensity is given by

The light intensity within this spot is 
t = 0.527 s
<u>It accelerates for 0.527 s.</u>
<u>Explanation:</u>
We use the formula:
v = u+at
Given:
v = 106 m/s
u = 0 (since no gravity)

So applying the formula,
v = u+at
106 = 0 + 201t
t = 106/201
t = 0.527 s