Which formula gives the maximum height of a projectile launched from ground level with speed v and angle θ?

• A. h = v^2 sin θ / (2g)
• B. h = v^2 cos^2 θ / (2g)
• C. h = v^2 sin^2 θ / g
• D. h = v^2 sin^2 θ / (2g)

Answer: (D)

The maximum height is determined by the vertical part of the motion, because gravity only acts to slow the upward vertical speed. At the top of the trajectory, the vertical velocity becomes zero, and all the initial vertical kinetic energy has been converted into gravitational potential energy.

Using v_y^2 = v_y0^2 - 2 g h with upward as positive, the final vertical velocity is zero at the highest point, and the initial vertical velocity is v sin θ. So 0 = (v sin θ)^2 - 2 g h, which gives h = (v^2 sin^2 θ) / (2 g). This is also what you get from energy conservation: (1/2) m (v sin θ)^2 = m g h.

So the maximum height depends on the square of the vertical speed component, leading to h = v^2 sin^2 θ / (2 g). The other expressions either miss the sin^2 factor, or place 2 g incorrectly, or involve the horizontal component, which doesn’t affect how high the projectile goes.