Answer:
[tex]t=2[/tex]
[tex]|v|=24.38693565m/s\\\alpha =34.54737228$^{\circ}$[/tex]
[tex]t=5[/tex]
[tex]|v|=25.41469214m/s\\\alpha =-37.78183103$^{\circ}$[/tex]
Explanation:
Let's use projectile motion equations, but first lets find the components of the initial speed vector:
[tex]v_ox=v_o*cos(\alpha _o)=39*cos(59)=20.08648492m/s[/tex]
[tex]v_oy=v_o*sin(\alpha _o)=39*sin(59)=33.42952473m/s[/tex]
Now let's find the magnitude and the angle of the velocity at t=2s, asumming g=9.8.
[tex]v_x=v_ox=20.08648492m/s[/tex]
[tex]v_y=v_oy-gt=33.42952473-(9.8*2)=13.82952473m/s[/tex]
Now the magnitude is given by:
[tex]|v|=\sqrt{(v_x)^{2} +(v_y)^{2}} =24.38693565m/s[/tex]
and its angle:
[tex]\alpha =arctan(\frac{v_y}{v_x} )=34.54737228$^{\circ}$[/tex]
Similarlyfor t=5
[tex]v_x=v_ox=20.08648492m/s[/tex]
[tex]v_y=v_oy-gt=33.42952473-(9.8*5)=-15.57047527m/s[/tex]
[tex]|v|=\sqrt{(v_x)^{2} +(v_y)^{2}} =25.41469214m/s[/tex]
[tex]\alpha =arctan(\frac{v_y}{v_x} )=-37.78183103$^{\circ}$[/tex]
