How to calculate jump height having only time in air.
It is possible to calculate jump height when i have only time "in air"? Time starts when my shoes/feets lose contact to floor and stop when touch it again (approximately). Half of all time my body will rise up and half of time fall down. It is correct? Both movements accelerations will be earth acceleration ~9,8 m/s^2? Is my thinking ok? So after half of time my body will be in top point and my velocity will be 0 m/s. Next half of time i will fall down with acceleration ~9,8 m/s^2. For example: When total time of jump = 0.6s => rise up time = fall down time = 0.3s. s = (a * t^2) / 2 = 0.441m. Has this sense :D? s = (9.8 * 0.3^2) / 2 s = (9.8 * 0.09) / 2 =
The acceleration is g, pointing downwards. Calling y(t) the height at time t:
y''(t) = -g
y'(t) = -g t + c0
y(t) = -g t² / 2 + c0 t + c1
We can find c0 and c1 using the initial conditions:
y(0) = 0
y'(t1) = 0 where t1 = 0.3 seconds
Thus:
y(0) = c1 = 0
y'(t1) = -g t1 + c0 = 0
c0 = g t1
Substituting c1 = 0 and c0 = g t1:
y(t) = - g t² / 2 + g t1 t
= g t (t1 - t/2)
Finally we can find the height reached at time t1 = 0.3 seconds:
y(t1) = g t1 (t1 - t1 / 2)
= g t1² / 2
= 9.8 × 0.3² / 2
= 0.4 m
If you jump for 0.6 seconds, then you will jump at a height of 0.4 m.