Answer
$5.27^\circ, 84.73^\circ$
Work Step by Step
As we know, $S=ut+\dfrac{1}{2}gt^2$
Also, $s=0$ because initial and final heights or distance is same.
This means $0=400 \sin \theta t+\dfrac{1}{2}(-9.8)t^2$
and $t=\dfrac{400 \sin \theta}{4.9}$
Now next step is to find the range of the particle.
Range: $R=(\dfrac{400 \sin \theta}{4.9}) (400 \cos \theta)$
or, $\dfrac{160000 \sin \theta \cos \theta}{4.9}=3000$
or, $\dfrac{ 2\times 160000 \sin \theta \cos \theta}{2 \times 4.9}=3000$
or, $3000=\dfrac{ 80000 \sin 2\theta}{4.9}$
or, $\sin 2\theta \approx (0.54422)$
Therefore, $\theta=5.27^\circ, 84.73^\circ$
