A projectile is launched from ground level with an initial velocity of 40 m/s at an angle of 30° above the horizontal. Assuming g = 10 m/s^2 and neglecting air resistance, what is the maximum height reached by the projectile?
Answer and explanation
Correct answer: C. 20 m
The vertical component of initial velocity is v_0y = v_0 sin(30°) = 40 m/s 0.5 = 20 m/s. At maximum height, vertical velocity is zero: v_y^2 = v_0y^2 - 2gh_max => 0 = (20)^2 - 2(10)h_max => 400 = 20 * h_max, yielding h_max = 20 m.
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