Question

1) A student jumps 0.85 meters straight up, on level ground. (a) How long is the student in the air? (b) For how long is the student within 0.10 meters of the top of the jump? (c) What fraction of the time is the student in the top 0.10 meters of the jump?

Answer 1

the height h is

h(t) = vt-4.9t^2
So, the max height is achieved when t = v/9.8
So, we have

v(v/9.8)-4.9(v/9.8)^2 = 0.85
v = 4.08
So, the height is roughly

h(t) = 4t-4.9t^2

Now you can figure the other answers, I expect.

Answer 2

This is how I did h = 1/2gt^2
t = sqrt2h/g = sqrt2*0.85/9.8
so t= 0.41628
total time = 2t = 0.832sec
v^2 = v^2 +2gh
v^2 = 0+2*9.81*0.1
v = 1.4
v= v+at
1.4 = v = 0+ 9.81 *t
t = 0.142 secs
total time 2t = 0.284sec
so 0.284/0.832 = 0.391

Answer 3

does that makes sense to you ?