Answer:[tex]\theta =57.99\approx 58 ^{\circ}[/tex]
Explanation:
Given
wall is 10 m from student
8 m high
Therefore student should launch at an angle \theta such that its maximum height is 8 m and its range is 20 m
For maximum height(h)[tex]=\frac{u^2sin^2\theta }{2g}[/tex]
Range (R)[tex]=\frac{u^2sin2\theta }{g}[/tex]
[tex]8=\frac{12^2sin^2\theta }{2g}-----1[/tex]
[tex]20=\frac{12^2sin2\theta }{g}----2[/tex]
divide 1 & 2
[tex]\frac{20}{8}=\frac{2sin2\theta }{sin^2\theta }[/tex]
[tex]\frac{5}{2}=\frac{2\times 2\times sin\theat \cdot cos\theta }{sin^2\theta }[/tex]
[tex]tan\theta =\frac{8}{5}[/tex]
[tex]\theta =57.99\approx 58 ^{\circ}[/tex]
