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A rugby player passes the ball 8.75 m across the field, where it is caught at the same height as it left his hand. At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used

Answer :

hamzaahmeds

Answer:

θ = 18.3°

Explanation:

The angle can be found by using the formula for the range of projectile motion:

[tex]R = \frac{u^2Sin2\theta}{g}[/tex]

where,

R = Horizontal Range = 8.75 m

u = launch speed = 12 m/s

g = acceleration due to gravity = 9.81 m/s²

θ = launch angle = ?

Therefore,

[tex]8.75\ m = \frac{(12\ m/s)^2\ Sin2\theta}{9.81\ m/s^2} \\\\Sin2\theta = \frac{(8.75\ m)(9.81\ m/s^2)}{(12\ m/s)^2} \\\\2\theta = Sin^{-1}(0.6)\\\theta = \frac{36.6^o}{2}[/tex]

θ = 18.3°

Since this angle is less than 45°. Therefore, it is the smaller of the two possible angles an dit is the correct answer:

θ = 18.3°

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