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?
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.
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
does that makes sense to you ?
To find the answers to these questions, we can use the equations of motion for free fall.
(a) To find the time the student is in the air, we can use the equation:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Rearranging the equation, we get:
t = sqrt(2h/g)
Substituting the given height h = 0.85 meters and the value of g, we can calculate the time the student is in the air.
(b) To find the time the student is within 0.10 meters of the top of the jump, we need to calculate the time it takes for the student to go from 0.85 meters to 0.75 meters (0.85 - 0.10). Using the same equation as before, we substitute h = 0.10 meters and solve for t.
(c) To find the fraction of time the student is in the top 0.10 meters of the jump, we can calculate the ratio of the time the student is within 0.10 meters of the top of the jump (from part b) to the total time in the air (from part a).
Now, you can plug in the values into the equations and calculate the answers.