Answer
(a) $\approx 3535m$ (b) $\approx 1531m$ (c) 200 m/s
Work Step by Step
(a) Given: $v_0=200 m/s$ and $\theta =60^\circ$
Since, we have $v(t)=\lt 200 \cos 60^\circ, 200 \sin 60^\circ-9.8 t\gt$
This implies that $v(t)=\lt 100 ,100 \sqrt3-9.8 t\gt$
As $r(t)=\int v(t) dt=\int [\lt 100 ,100 \sqrt3-9.8 t\gt]dt=\lt 100t ,100 \sqrt3t-4.9t^2\gt$
Range will be: $t=\dfrac{2 (200)(sin 60)}{g} \approx 3535m$
(b) From part (a), we have $t=35.35$
Thus, Maximum height$=100 \sqrt3(17.58)-4.9(17.68)^2=\approx 1531 m$
(c) Speed is given by $s(t)=|v(t)|=\sqrt{(100)^2+(100 \sqrt3-9.8 t)^2}=200 m/s$
Hence, the required answers are:
(a) $\approx 3535m$ (b) $\approx 1531m$ (c) 200 m/s
