ok so here is what I have done:
s = vt - 1/2at^2 -- I have never used this formula before, found it off google
s = ymax
a= -9.81 m/s^2
t = 0.5 (because that is half the total time to go up and down and I am just finding the up)
v = 0
ymax = 9.81 x 1/2 x 0.5^2 = 1.23m, so ymax/2 =...
Ok I think I've calculated t1, but I don't think it is correct.. if you are doing a calculation for an object being thrown in the air, until it reaches the point where it is coming back down can you set v=0?
In the vertical jump, an athlete starts from a crouch and jumps upward to reach as high as possible. Even the best athletes spend little more than 1.00 s in the air (their "hang time"). Treat the athlete as a particle and let ymax be his maximum height above the floor.
To explain why he seems...
You are climbing in the High Sierra where you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top and 8.50s later hear the sound of it hitting the ground at the foot of the cliff.
Ignoring air resistance, how high is the...