Question

Un proiectil lansat cu viteza vo=240 m/s sub unghiul alfa=60° fata de orizontala loveste varful unui deal de inaltime h=500 m. Care este distanta orizontala pana la tinta si durata miscarii proiectilului?​

Answer 1

Ox:x=v₀cosαt

Oy:y=v₀sinαt-gt²/2

voy=v₀sinα

vox=v₀cosα

y=h=>h=v₀sinαt-gt²/2<=>500=240*0,866t-5t²<=>500=207,84t-5t²<=>5t²-207,84t+500=0 |:5=>t²-41,56+100=0

Δ=1727,89-400=1327,89

t₁,₂=(41,56±36,44)/2

t₁=(41,56+36,44)/2=78/2=39 s

t₂=(41,56-36,44)/2=5,12/2=2,56 s

t₂<<

t=t₁=39 s

d=v₀cosαt=240*0,5*39=4680 m