A hobbyist launches a projectile from ground level on a horizontal plain.It reaches a maximum height of 72.3 m and lands 111 m from the launch point,with no appreciable air resistance.What was the angle of launch if g = 9.80 m/s²? A)69.0° B)67.4° C)22.6° D)44.8°

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Quizplus · Accepted Answer

The correct answer is A because the projectile reaches its maximum height when the vertical component of its velocity is zero. The horizontal component of its velocity is constant throughout its flight. The range of the projectile is the horizontal distance it travels before landing. The angle of launch can be calculated using the following equation:$$	heta = 	an^{-1}\left(\frac{v_i^2 \sin 2	heta}{gR}ight)$$where:$$	heta$$ is the angle of launch$$v_i$$ is the initial velocity of the projectile$$g$$ is the acceleration due to gravity$$R$$ is the range of the projectileSubstituting the given values into the equation, we get:$$	heta = 	an^{-1}\left(\frac{(111 	ext{ m})^2 \sin 2	heta}{(9.80 	ext{ m/s}^2)(72.3 	ext{ m})}ight)$$Solving for $$	heta$$, we get:$$	heta = 69.0^\circ$$

A Hobbyist Launches a Projectile from Ground Level on a Horizontal