Math
posted by Conder on .
The highest dive in the Olympic Games is from a 10meter platform. The height "h" is in meters of a diver "t" seconds after leaving the platform can be estimated by the equation, h=10+4.9t4.9t^2
I don't know how to make a successful table for this. I tried 0.1, 0.2, etc.. but that doesn't seem to work.

when t=.1 , h = 10 + 4.9(.1)  4.9(.1)^2 =10.441
when t= .2 , h= 10 + 4.9(.2)  4.9(.2)^2 = 10.784
what do you mean it "doesn't seem to work" ?
He clearly jumps upwards for a short time before his height decreases.
Just keep going the way I showed you. 
The general equation for the height of an object falling from a height of H is
h = H  Vot  9.8t^2/2 where H = the initial height in meters, Vo = the initial downward velocity in m/sec., 9.8 = the acceleration due to gravity in m/sec.^2 and t = the time to impact in seconds.
Your equation should therefore read
h=104.9t4.9t^2, where I assume you have selected H = 10 and Vo as 4.9 m/s.
For t = .10, h = 10  .49  .049 = 9.461m.
For t = .20, h = 10  .98  .196 = 8.824m.
For t = .30, h = 10  1.47  .441 = 8.089m.
t.10.20.301.973
h...9.461..8.824..8.089.....10