A stone is projected horizontally with a speed of 15\text{ m/s} from the top of a vertical… — JAMB UTME

A stone is projected horizontally with a speed of $15\text{ m/s}$ from the top of a vertical cliff of height $20\text{ m}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the magnitude of the velocity of the stone just before it strikes the ground?

A
$15\text{ m/s}$
B
$20\text{ m/s}$
$25\text{ m/s}$ Answer
D
$35\text{ m/s}$

Answer

The magnitude of the velocity of the stone just before hitting the ground is

25\text{ m/s}

For a horizontally launched projectile, the horizontal velocity component remains constant at

v_x = 15\text{ m/s}

. The vertical component just before impact is found using

v_y^2 = u_y^2 + 2gh = 0 + 2(10)(20) = 400

, giving

v_y = 20\text{ m/s}

. Combining these perpendicular components yields a total speed of

v = \sqrt{15^2 + 20^2} = 25\text{ m/s}

Step-by-Step Solution

Identify horizontal velocity component

v_x = 15\text{ m/s}

Air resistance is neglected, so horizontal velocity remains constant throughout flight.

Calculate vertical velocity component just before impact using third equation of motion

v_y^2 = u_y^2 + 2gh = 0^2 + 2(10)(20) = 400 \implies v_y = 20\text{ m/s}

Initial vertical velocity

u_y = 0\text{ m/s}

for horizontal projection; stone falls through a vertical displacement of

20\text{ m}

under gravity.

Compute total resultant velocity magnitude using Pythagorean theorem

v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ m/s}

Horizontal and vertical velocity components are mutually perpendicular.

Key Concept

Horizontal Projection and Resultant Velocity Vector Synthesis

Estimated Time:2m 0s

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