A projectile is launched from ground level at speed 15 m/s at 30 degrees. What is its maximum height neglecting air resistance?

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Multiple Choice

A projectile is launched from ground level at speed 15 m/s at 30 degrees. What is its maximum height neglecting air resistance?

Explanation:
The maximum height is set by the vertical motion under gravity, with the vertical speed reaching zero at the top. The vertical component of the initial velocity is v0y = v0 sin(30°) = 15 × 0.5 = 7.5 m/s. Using the kinematic relation v_y^2 = v0y^2 − 2 g h and setting v_y = 0 at the top gives h = v0y^2 / (2 g) = (7.5)^2 / (2 × 9.8) = 56.25 / 19.6 ≈ 2.87 m. This matches the provided value. (Equivalently, H = v0^2 sin^2 theta / (2 g) also yields the same result.) Since air resistance is neglected, gravity alone determines the height.

The maximum height is set by the vertical motion under gravity, with the vertical speed reaching zero at the top. The vertical component of the initial velocity is v0y = v0 sin(30°) = 15 × 0.5 = 7.5 m/s. Using the kinematic relation v_y^2 = v0y^2 − 2 g h and setting v_y = 0 at the top gives h = v0y^2 / (2 g) = (7.5)^2 / (2 × 9.8) = 56.25 / 19.6 ≈ 2.87 m. This matches the provided value. (Equivalently, H = v0^2 sin^2 theta / (2 g) also yields the same result.) Since air resistance is neglected, gravity alone determines the height.

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