V^2=u^2 +2aS
U is found first by considering that first 8 secs and using v=u+at. {different v and u though}
V=-u+gt.
Magnitude of u = magnitude of v if there is no resistance ( because the conservation of energy says the k. E. must be the same when it passes you as when it left your hand).... up is negative here, down is positive.
V+v=gt
2v= g x 8
V=4xg.= the initial velocity for the next calculation
V^2=(4g)^2+(2xgx21)
So v can be calculated.
The height of the ball above the ground is 38.45 m
First we will calculate the velocity of the ball when it touch the ground by using first equation of motion
v=u+gt
v=0+9.81×2.8
v=27.468 m/s
now the height of the ground can be calculated by the formula
v=√2gh
27.468=√2×9.81×h
h=38.45 m
Answer:
Nuclease is the answer I know
I hope this is the answer
Because they both have to do with chemistry
Answer:
At light intensity I = 3, is P a maximum
Explanation:
Given:

now differentiating the above equation with respect to Intensity 'I' we get

or

or

or

Now for the maxima 
thus,

or

or

or

or
I = 3
thus, <u>for the value of intensity I = 3, the P is maximum</u>
at I = 3

or

or
