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The range of a projectile that is launched with an initial velocity v at an angle of a with the horizontal is given by R

sin

where g is the acceleration due to gravity or 9.8 meters per second squared. If a projectile is launched with an initial velocity of 1

meters per second, what angle is required to achieve a range of 20 meters? Round answers to the nearest whole number.

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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