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Physics 22 Online
OpenStudy (anonymous):

An object is dropped from a height of 100m, How long is it in the air?

OpenStudy (anonymous):

Assuming no air resistance,\[height(t)=h_0-\frac{1}{2}g t^2\] We want to find the time when h(t)=0 \[0=100-4.9 t^2\] \[4.9 t^2=100\] \[ t=\sqrt{100/4.9}\]

OpenStudy (anonymous):

initial velocity Vi=0 m/s Height= h= 100m and g=9.8m/s^2 time=T=? using second equation of motion.. h=ViT+(gT^2)/2 h= (0)(T)+ (9.8/2) T^2 h= 4.9 T^2 100=4.9 T^2 T^2=100/4.9 T^2= 20.41 T= sqrt (20.41) T= 4.52 sec

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