Answer: You throw an object upwards from the ground with a velocity of 20m/s and it is subject to a downward
acceleration of 9.8 m/s2. How high does it go?
Given Formula Set up Solution
vi=20 m/s
vf=0 m/s
a=-9.8 m/s2
x=?
vf
2 = vi
2 + 2a x (0 m/s)2= (20 m/s)2 + 2 (-
9.8 m/s2) x
x=20.4 meters
Explanation: i known/ i hoped that helped.
Answer:
Explanation:
Given
height of building h=3 m
Landing velocity of diver
at an angle of 
Let u be the initial velocity of diver at an angle of \theta with horizontal
Since there is no acceleration in horizontal direction therefore horizontal component of velocity will remain same
---- -----1
Considering Vertical motion

here 




----------------2
Divide 2 and 1 we get



Thus 

One point will be X1,Y1 and the other will be X2,Y2. It does not matter which is which except that X1 and Y1 have to be the same point and X2 and Y2 have to be the same point. For example, let's say you were given (2,3) and (6,8). No matter which point is X1,Y1 and the other is X2,Y2, the slope will still be 5/4.
The rise is the change in y from one point to the other. The run would be the change in x from one point to the other.
Answer:
Option (B)
Explanation:
In the stabilizing natural selection, the extreme traits from both the ends are eliminated by natural selection and natural selection favors the intermediate trait. So over time individuals having the intermediate traits are selected over the individuals having extreme traits.
So here the population of the bat which possesses moderate wing length is selected over the individual with extreme traits like individuals with short wings and long wings. As a result, the population of moderate length wing bats increased.
Therefore the correct answer is (B)- stabilizing natural selection.