Answer
At $t=0.8~s$, the magnitude of the velocity will be 100 m/s
Work Step by Step
The east component of the velocity vector will continue to be 60 m/s as the north component of the velocity vector increases. Let $v_n$ be the north component of the velocity vector when the magnitude of the velocity is 100 m/s. We can find $v_n$:
$\sqrt{v_n^2+(60~m/s)^2} = 100~m/s$
$v_n^2+(60~m/s)^2 = (100~m/s)^2$
$v_n^2 = (100~m/s)^2 - (60~m/s)^2$
$v_n = \sqrt{(100~m/s)^2 - (60~m/s)^2}$
$v_n = 80~m/s$
We can find the time when the north component of velocity is 80 m/s:
$v_f = v_0+at$
$t = \frac{v_f-v_0}{a}$
$t = \frac{80~m/s-0}{100~m/s^2}$
$t = 0.8~s$
At $t=0.8~s$, the magnitude of the velocity will be 100 m/s
