Answer:
The angle the projectile was fired is [tex]45^o[/tex]
Explanation:
Recall the formulas for maximum height and ranges for a projectile fired with initial velocity "v" at an angle [tex]\theta[/tex]:
[tex]h = \frac{v^2\,sin^2(\theta)}{2\,g}\\R=\frac{v^2\,sin(2\,\theta)}{g}[/tex]
we can use them to solve for the angle by first isolating the value [tex]v^2[/tex] which is common in both equations:
[tex]v^2=2\,h\,g/sin^2(\theta)=2\,(15)\,g/sin^2(\theta)=30\,(g)/sin^2(\theta) \\ \\v^2=R\,g/sin(2\,\theta)=60\,(g)/sin(2\,\theta)[/tex]
and now, making those to expressions equal and using the formula for the sine of a double angle, we get:
[tex]\frac{30\,(9.8)}{sin^2(\theta)} =\frac{60\,(g)}{sin(2\,\theta)} \\30\,(g)\,sin(2\,\theta)=60\,(g)\,sin^2(\theta)\\sin(2\,\theta)=2\,sin^2(\theta)\\2\,sin(\theta)\,cos(\theta)=2\,sin(\theta)\,sin(\theta)\\cos(\theta)=sin(\theta)[/tex]
This happens when [tex]\theta=45^o[/tex]
