Asked by Anonymous
A window is 1.50m high. A stone falling from above passes the top of the window with a speed of 3m/s. When will it pass the bottom of the window?
Answers
Answered by
Martin
x = x0 + v0*t + 1/2*a*t^2
x0=1.5 v0=-3 a=-10 (or something close, depending on what you're told to use in your class).
You want the time for when x=0 (where 0m = the bottom of the window).
0 = 1.5 - 3 x t - 5 x t^2
1.5 = 3xt + 5xt^2
0=5t^2 +3t -1.5
You use the quadratic equation and you find : t ~= 0.3245
x0=1.5 v0=-3 a=-10 (or something close, depending on what you're told to use in your class).
You want the time for when x=0 (where 0m = the bottom of the window).
0 = 1.5 - 3 x t - 5 x t^2
1.5 = 3xt + 5xt^2
0=5t^2 +3t -1.5
You use the quadratic equation and you find : t ~= 0.3245
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