A particle is projected horizontally from the top of a tower with speed 10 m/s. Neglect air resistance and take g = 10 m/s². At the instant its velocity makes an angle 60° below the horizontal, its vertical displacement below the point of projection is:
Options

Question

A particle is projected horizontally from the top of a tower with speed 10 m/s. Neglect air resistance and take g = 10 m/s². At the instant its velocity makes an angle 60° below the horizontal, its vertical displacement below the point of projection is:

Options

G1 Physics · Accepted Answer

Horizontal velocity is 10 m/s. At 60° below horizontal, tan60° = vᵧ/vₓ, so vᵧ = 10√3 m/s downward. Since vᵧ = gt, t = √3 s. Vertical displacement = (1/2)gt² = 5×3 = 15 m. Therefore option C is correct.