KinematicsMedium

Stone thrown from a bridge

A stone is thrown horizontally from a bridge h = 45 m high with initial velocity v₀ = 15 m/s.\nCalculate: (a) fall time, (b) horizontal range, (c) impact velocity.

Given data

h = 45 mv₀ = 15 m/s (horizontal)g = 9.81 m/s²v₀y = 0 (horizontal throw)

Review the theory: Cinematica

Steps

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How long does the stone take to fall from h = 45 m? (time in seconds)
Write your working — the numerical result must appear in the text s
What is the horizontal distance travelled before impact?
What is the vertical velocity at the moment of impact?
Calculate the total velocity at impact (magnitude of the velocity vector).

Full worked solution

How long does the stone take to fall from h = 45 m? (time in seconds)
$t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \cdot 45}{9.81}}$
$t = \sqrt{2h/g} = \sqrt{2\cdot45/9.81} = \sqrt{90/9.81} = \sqrt{9.174} \approx \mathbf{3.03\,s}$ . Vertical motion is free fall from rest, independent of the horizontal velocity $v_0$ .
What is the horizontal distance travelled before impact?
$x = v_0 \cdot t = 15 \cdot 3.03$
$x = v_0 \cdot t = 15 \times 3.03 \approx \mathbf{45.4\,m}$ . Horizontal motion is uniform with constant velocity — no acceleration in the $x$ direction.
What is the vertical velocity at the moment of impact?
$v_y = g \cdot t = 9.81 \cdot 3.03$
$v_y = g\,t = 9.81 \times 3.03 \approx \mathbf{29.7\,m/s}$ downward. Vertical velocity increases linearly from zero due to constant gravitational acceleration.
Calculate the total velocity at impact (magnitude of the velocity vector).
$v = \sqrt{v_0^2 + v_y^2} = \sqrt{15^2 + 29.7^2}$
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 29.7^2} = \sqrt{225 + 882} = \sqrt{1107} \approx \mathbf{33.3\,m/s}$ . Direction: $\theta = \arctan(v_y/v_x) = \arctan(29.7/15) \approx \mathbf{63^\circ}$ below the horizontal.

Result:Fall time: 3.03 s — Range: 45.4 m — Impact velocity: 33.3 m/s at 63° below horizontal.

Stone thrown from a bridge

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