Zadanie w załączniku :) Proszę o rozwiązanie dwóch ostatnich zadań. Dzięki ! :D

Zad 3. [latex]Ek=frac{mv^2}{2}Rightarrow v=sqrt{frac{2Ek}{m}}\\ v=sqrt{frac{2*320}{0,4}}=40frac{m}{s}[/latex] [latex]z=frac{v^2sin2alpha}{g}Rightarrow alpha=dfrac{arcsin(frac{zg}{v^2})}{2}\\ alpha=dfrac{arcsin(frac{50*10}{40^2})}{2}\\ alpha=9^o[/latex] [latex]h_{m}=frac{(v sinalpha)^2}{2g}\\ h_m=frac{(40*0,1564)^2}{2*10}\\ h_m=1,96m[/latex] [latex]Ep_{max}=mgh_m\\ Ep_{max}=0,4*10*1,96\\ Ep_{max}=7,84J[/latex] Zad 4. [latex]x(t)=Asin(omega t)\\ v(t)=frac{dx}{dt}=omega Acos(omega t)\\ a(t)=frac{dv}{dt}=-omega^2Asin(omega t)\\ v^2=omega^2 A^2cos^2(omega t)\\ v^2=omega^2 A^2[1-sin^2 (omega t)]\\ v^2=omega ^2A^2-A^2sin^2(omega t)\\ v^2=omega ^2A^2-x^2Rightarrow A=sqrt{frac{v^2+x^2}{omega^2}}[/latex] [latex]A=sqrt{frac{15^2+4^2}{3^2}}\\ A=5,2cm[/latex] [latex]x(t=2s)=5,2sin(3*2)\\ x(t=2s)=5,2*(-0,28)\\ x(t=2s)=-1,456cm[/latex] Pozdrawiam, Adam