A projectile is launched from ground level at an angle of 30 degrees above the horizontal with initial speed 40 m/s. Assuming g = 9.8 m/s^2 and neglecting air resistance, what is its maximum height above the ground?
Answer and explanation
Correct answer: D. 20.4 m
- Step 1: Find the initial vertical velocity component: v_{0y} = v_0 sin(30 degrees) = 40 0.5 = 20 m/s.
- Step 2: At maximum height, the vertical velocity is zero: v_y = 0.
- Step 3: Use kinematic equation v_y^2 = v_{0y}^2 - 2gh: 0 = 20^2 - 2 9.8 h.
- Step 4: Solve for h: h = 400 / (19.6) = 20.41 m.
Answer
20.4 m.
Why not the others: 40.8 m computes 400/9.8 omitting the factor of 2; 15.3 m uses cos(30 degrees); 81.6 m uses the total launch speed 40 m/s without resolving components.
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