Recollect the past 5!

$\text{Let the initial velocity be }x \text{ and the angle be }\theta.\\\text{Then, the }y\mbox{ and }x \text{ coordinate are given by: }\\y=v\sin(\theta)t-5t^2\\x=v\cos(\theta)t\\\text{Combining the two, we have}\\y=\tan(\theta)x-\frac{5}{v^2\cos^2(\theta)}x^2\\\text{}\\\text{Referring to the values given in the question, we have}\\10=10\tan\theta-\frac{500}{v^2\cos^2\theta}=20\tan\theta-\frac{2000}{v^2\cos^2\theta}\\\text{Utilising the last equation, we derive the following:}\\v=\sqrt{\frac{150}{\cos^2\theta\tan\theta}}\\10=\frac{20}{3}\tan\theta\\\text{From here, it follows that }v=18.03$

From simple analysis of the situation we say range to be 3a i.e. 30m = u^2 sin(2theta)/g And from y=x tan(theta)[(1)-(x/R)] we get tan of theta to be 3/2. This by putting in equation one gives answer to be 18.02 m/s