Question

Help please, show work please :D

You have fallen from a height of 9.09

1. How long were you falling?
2.What was your velocity just before you hit the ground?
3. Your fall was broken in 0.00701s
What was your braking distance?

Answer 1

s = displacement
u = initial velocity
t = time
a = acceleration\[s = ut+\frac{at^2}{2}\]Therefore\[(t+\frac{u}{a})^2=\frac{2s}{a}+\frac{u^2}{a^2}\]Therefore\[t=\frac{-u+\sqrt{u^2+2as}}{a}\]I dropped the minus from the plus or minus in front of the square root because we're looking for the positive value of t.
Now just mash the right keys on your calculator and part 1. is sorted!

Answer 2

v = final velocity
u = initial velocity
a = acceleration
s = displacement\[v^2=u^2+2as\]Now just put the right numbers in, square root both sides and part 2. is done!

Answer 3

Assuming that, during braking, average velocity was half of initial velocity then the braking distance is\[\frac{0.00701v}{2}m\]Otherwise, obtain an expression for velocity with respect to time and integrate that between the start and end times of the braking.

Answer 4

Thanks CheesePie !! :D