Dave Above Ground

By · Monday, August 31st, 2009

Dave Above Ground
Please solve showing work?

For 1989 and 1990, Dave Johnson had the highest score in the decathlon world.When Johnson reached a speed of 32 m / sec in the cault pole of the runway, the height above ground h. t seconds after leaving the land was given by h =- 162 ^ 2 +32 t. How time was over 14 feet in the air? Please show your work

I think your equation should read "h =- 16t ^ 2 + 32t – the -162 "^ 2" part is meaningless. He goes up and down following a parable, so it must be ^ 2 in the term in any part of this equation for the work. I will assume to be-16t ^ 2 and go from there. To find the point where it reaches 14 feet, simply set the equation to 14: h = 16t ^ 2 + 32t = 14-16t ^ 2 + 32t 0 =- 16t ^ 2 + 32t – 14 0 = 8t ^ 2 – 16t + 7 t = [16 + / – sqrt ((-16) ^ 2 – 4 (8) (7))] / 2 ( 8) t = [16 + / – sqrt (256 – 224)] / 16 t = [16 + / – sqrt (32)] / 16 t = 1 + / – sqrt (2) / 4 t = 0.65 s, 1.35s Thus, 0.65sy 1.35s, was at 14 feet. Subtract the two and you find that was at or above 14 feet in 0.7s. (For a precise time, you can subtract [1 + sqrt (2) / 4] – [1 – sqrt (2) / 4] and get sqrt (2) / 2.)

Amazing Cicada life cycle – Sir David Attenborough’s Life in the Undergrowth – BBC wildlife

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