Question

?

Answer 1

horizontal distance is not relevant here since we want to find the time it takes to cross a vertical distance only

use kinematics equation 3 to solve for t

Answer 2

x=0.35+1/2(9.8)(2.1)

Answer 3

delta x is the distance

we are solving for time not distance

Answer 4

0.35=d+1/2(9.8(2.1)

Answer 5

the vertical initial velocity is 0

Answer 6

ugh

Answer 7

so set it equal to 0

Answer 8

v0 needs to be 0 yes

Answer 9

0=d+1/2(9.8(2.1)

Answer 10

the equation is

deltax = v0 + (1/2)at^2

Answer 11

substitute deltax, v0, a, then solve for t

Answer 12

deltax=0+(1/2)(2.1)t

Answer 13

a is the acceleration due to gravity

again, horizontal velocity is not relevant here

Answer 14

also it is t^2 not just t

Answer 15

0+(1/2)(9.8)t^2

Answer 16

good, solve for t

Answer 17

wait what is it equal to 0?

Answer 18

deltax = ?

Answer 19

the disrance

Answer 20

so 0.35

Answer 21

good, now solve for t

Answer 22

0.3

Answer 23

B

Answer 24

let's try to keep a few more sig figs (so less rounding)

t = 0.26 = A

Answer 25

0.27 really