The maximum displacement angle is given as 13°
L = length = 1.25 meters
Initial velocity = 0.8m/s
displacement = ∅max
Work done is given as change in kinetic energy
[tex]\frac{1}{2} mvi^2+0=0+mgh\\\\h = \frac{vi^2}{2g} \\\\h = \frac{0.8^2}{2*10} =\frac{0.64}{20} =0.032 meteres[/tex]
L-h
= 1.25-0.032
= 1.218 meters
Cos∅max = x/L
[tex]=\frac{1.218}{1.25} \\\\= 0.9744\\\\[/tex]
∅max = Cos⁻¹(0.9744)
= 12.99°
≈ 13°
Therefore the maximum displacement angle is given as 13°
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