A stone is projected horizontally with a velocity 9.8 ms^–1 from a tower of height 100 m. Its velocity one second after projection is

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9.8 ms^-1

4.9 ms^-1

9.8√2 ms^-1

4.9√2 ms^-1

answer is C.

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Detailed Solution

We are asked to determine the velocity of a stone projected horizontally with speed u_x = 9.8 m s⁻¹ from a tower of height 100 m, at a time t = 1 s after projection. Air resistance is neglected.

Horizontal motion: Because there is no horizontal acceleration, the horizontal velocity remains constant: v_x = 9.8 m s⁻¹.
Vertical motion: The initial vertical velocity is zero. Under gravity g = 9.8 m s⁻², the vertical velocity after time t is v_y = g t. At t = 1 s, v_y = 9.8 m s⁻¹ downward.
Resultant speed: The total velocity magnitude is the vector sum of the components: v = √(v_x² + v_y²) = √(9.8² + 9.8²) = 9.8√2 m s⁻¹.