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Katenokk
@Katenokk
September 2021
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При свободном падении на пути 35 м скорость тела увеличилась до 40 м/с. Определите скорость тела в начале пути.
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Гоша68
Verified answer
S=v0t+gt^2/2
v=v0+gt
35=v0t+10t^2/2 v0t+5t^2=35
40=v0+10t v0+10t=40 v0=40-10t
40t-10t^2+5t^2=35 5t^2-40t+35=0
t^2-8t+7=0
t1=1 t2=7
v0=40-10=30 cкорость в начале 30 м/c
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Answers & Comments
Verified answer
S=v0t+gt^2/2v=v0+gt
35=v0t+10t^2/2 v0t+5t^2=35
40=v0+10t v0+10t=40 v0=40-10t
40t-10t^2+5t^2=35 5t^2-40t+35=0
t^2-8t+7=0
t1=1 t2=7
v0=40-10=30 cкорость в начале 30 м/c