White sitting on a tree branch 10.0m above the ground, you drop a chestnut. When the chestnut has fallen 2.5m, you throw a second chestnut straight down. (A) What velocity does the first chestnut have after it has dropped 2.5m? (B) How long does it take the first chestnut to fall the last 7.5m? (C) What initial speed must you give the second chestnut if they are both to reach the ground at the same time?

Explanation:

A )

v² = u² + 2 a s

u = 0

s = 2.5

a = 9.8

v² = 2 x 9.8 x 2.5 = 49

v = 7 m /s

B )

time required to fall by first 2.5 m , t₁

s = 1/2 a t₁²

2.5 = .5 x 9.8 t₁²

t₁ = .714 s

time required to fall by 10 m

s = 1/2 a t₂²

10 = .5 x 9.8 t₂²

t₂ = 1.43 s

time required to fall by last 7.5 m

= t₂ - t₁

= 1.43 - .714

= .72 s

C )

time taken by second = .72 s  

For second let initial velocity required be u

s = ut + 1/2 g t²

10 = u x .72 + .5 x 9.8 x .72²

10 = u x .72 + 2.54

u = 10.36 m /s

double the distance = divide strength by 4

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.