Rwoanmuhammedrans
Answered

Yerden dikey yönde 5 m/s hızla fırlatılan bir taş, mak- simum yüksekliğin üçte ikisindeyken (2ymax/3) hızı ne olur? [C: 4.1 m/s.] ​

Sagot :

Cevap:

hmax=V²/2.g=5²/2.10=25/20=5/4  m

h1=5/4 .2/3=10/12=5/6 m

V²=Vo²-2gh

V²=5²-2.10.5/6

V²=25-50/3=25/3

V=5/√3=(5√3)/3

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