Physics: Projectiles

A person has a vertical leap of 1.0 m. Not too bad! How much time is this person in the air? With what velocity does this person leave the ground?

Where do I start? What formula should I use? Help Please.

Well. The formula is vf^2 = vi^2 + 2a(df - di). We know the final velocity is zero at the highest point, so we will set vf to zero. We are solving for initial velocity, acceleration by gravity is -9.8 meters per second, the distance is 1m. So 0 = vi^2 + 2(-9.8)(1). 0 = vi^2 + (-19.6). 19.6. = vi^2. Finally take the square root of 19.6 to find the answer of 4.427 meters per second. Then you can plug that number into this equation: tf = change in velocity divided by the constant acceleration. Tf = 4.4/9.8. Which is equal to .903 seconds.

Conservation of mechanical energy.

Emechanical = Uinitial + Kinitial = Ufinal + Kfinal
U is potential energy (mgh), and K is kinetic energy ((1/2)mv^2)
Uinitial is 0 because h is 0.
Kfinal is 0 because v is 0.
0 + Kinitial = Ufinal + 0
Kinitial = Ufinal
(1/2)mv^2 = mgh
Cancel out the masses...
(1/2)v^2 = gh
h = 1 from the problem statement.
(1/2)v^2 = 1g
g is 9.81
(1/2)v^2 = 9.81
v^2 = 19.62
v = sqrt(19.62)
v = 4.42945 m/s

So that solves the second part of the problem - the initial velocity is 4.42945 m/s - now work backwards to get the time in the air.

xf = x0 + vt + (1/2)at^2
xf and x0 are both 0 at the ground...
0 = 0 + vt + (1/2)at^2
Get v from above, and a is -9.81 due to gravity.
0 = (4.42945)t - (1/2)(9.81)t^2
Factor that out...
0 = t(4.42945 - (1/2)(9.81)t)
0 = t(4.42945 - 4.905t)
t = 0.903048s

Peace.