Physics Problem?

Question

A common flea is recorded to have jumped as high as 21 cm. Assuming that the jump is entirely in the vertical direction and that air resistance is insignificant, calculate the time it takes the flea to reach a height of 7.0 cm.

the answer is 0.04 s. i know how to get it, but i don't get why.
so i plugged it into that equation:
delta y = (initial velocity)(delta T) + (0.5)(acceleration)(time-squared)

0.07 = 0 + (0.5)(9.81)(time-squared)
t=0.12

so i divided 0.12 by 3. and i got 0.04 seconds.
why do you divide by 3? 7 cm is already a third of the highest jump of the common flea. wasn't i calculating the 7.0 cm jump time? not the 21 cm one?

or did i just completely fail at this problem?
help, thank you.

Answer 1

Unfortunately, your analysis is not correct. You did not even use the given data "A common flea is recorded to have jumped as high as 21 cm". Did you think that this piece of information is useless?
First, please calculate how long it would take the flea to reach 21cm:
0.21 = 0.5*9.81*T^2
or T = 0.207(s)
Remember: this formula s=0.5gt^2 can only be used for either the initial or the final velocity to be zero. However, neither the initial nor the final velocity to be zero for the jump to reach a height of 7.0 cm. Nevertheless, you may image that the flea continues to move past this 7.0cm and finally reach the height of 21cm. The time to take to complete the later part of the jump can be calculated:
(0.21 - 0.07) = 0.5*9.81*t^2
or t = 0.169(s)
The time difference of T - t is what you are looking for:
T - t = 0.207s - 0.169s = 0.038s ≈ 0.04s

In your analysis of:
delta y = (initial velocity)(delta T) + (0.5)(acceleration)(time^2)
0.07 = 0 + (0.5)(9.81)(time-squared)
Neither (initial velocity) nor (T) is zero, but you set them to be zero?