Answer
(a) $\approx 3535m$ (b) $\approx 1531m$ (c) 200 m/s
Work Step by Step
(a) As we are given that $v_0=200 m/s$ and $\theta =60^\circ$
From question: $v(t)=\lt 200 \cos 60^\circ, 200 \sin 60^\circ-9.8 t\gt$
or, $v(t)=\lt 100 ,100 \sqrt3-9.8 t\gt$
As $r(t)=\int v(t) dt$
or, $r(t)=\int [\lt 100 ,100 \sqrt3-9.8 t\gt]dt$
or, $r(t)=\lt 100t ,100 \sqrt3t-4.9t^2\gt$
Need to find the range.
$t=\dfrac{2 (200)(sin 60)}{g} \approx 3535m$
(b) As we are given that $v_0=200 m/s$ and $\theta =60^\circ$
Use part (a), we have $t=35.35$
Need to find the Maximum height$=100 \sqrt3(17.58)-4.9(17.68)^2$
or, $=\approx 1531 m$
(c) As we are given that $v_0=200 m/s$ and $\theta =60^\circ$
Speed is given by $s(t)=|v(t)|=\sqrt{(100)^2+(100 \sqrt3-9.8 t)^2}=200 m/s$
