Answer
3.43 m/s
Work Step by Step
Please see the attached image first.
Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction to find the vertical displacement.
$\downarrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$y=0+\frac{1}{2}(9.8\space m/s)^{2}(1.1\space s)^{2}=5.9\space m$
Given that the magnitude of the displacement vector of the projectile after 1.1 s = 7 m, So using the Pythagorean theorem we can write.
$(7\space m)^{2}=x^{2}+(5.9\space m)^{2}$
$x=3.77\space m$
Let's apply the equation $S=ut$ in the horizontal direction to find the initial velocity of the launch.
$\rightarrow S=ut$ ; Let's plug known values into this equation.
$3.77\space m=u\times 1.1\space s$
$u=3.43\space m/s$
