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Sagot :
Answer:
[tex]\theta=30.285^{\circ}[/tex]
Step-by-step explanation:
The range of a projectile is given by :
[tex]R=\dfrac{u^2\sin2\theta}{g}[/tex]
Put R = 20 m, u = 15 m/s and finding the value of angle of projection
So,
[tex]R=\dfrac{u^2\sin2\theta}{g}\\\\\sin2\theta=\dfrac{Rg}{u^2}\\\\\sin2\theta=\dfrac{20\times 9.8}{15^2}\\\\\sin2\theta=0.871\\\\2\theta=\sin^{-1}(0.871)\\\\2\theta=60.57\\\\\theta=30.285^{\circ}[/tex]
So, the required angle of projection is equal to [tex]30.285^{\circ}[/tex].
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