Acceleration, a = ( 4.80 i+7.00 j )[tex]m/s^2[/tex]
We have acceleration a = Horizontal component i + Vertical component j
So Horizontal component =4.80 [tex]m/s^2[/tex]
Vertical component =7.00 [tex]m/s^2[/tex]
We have horizontal displacement , [tex]s = ut+\frac{1}{2} at^2[/tex]
Here s = 12.4 m, u = 4.4 m/s and a = 4.80 [tex]m/s^2[/tex]
Substituting
[tex]12.4 = 4.4*t+\frac{1}{2} *4.80*t^2\\ \\ t^2+1.83*t-5.167=0\\ \\ (t-1.53)(t+3.36)=0\\ \\ time = 1.53 seconds[/tex]
Negative time is not possible.
Now we need to find velocity after 1.53 seconds
We have v = u+at
Horizontal case
v = 4.4 + 4.80 * 1.53 = 11.744 i
Vertical case
v = 0 + 7.00 * 1.53 = 10.71 j
So Velocity when it has been displaced by 12.4 m parallel to the x axis = (11.744 i+10.71 j) m/s
Magnitude = [tex]\sqrt{11.744^2+10.71^2} =15.89 m/s[/tex]
Angle , tan θ = [tex]\frac{10.71}{11.744}[/tex]
θ = 42.36⁰
