A body is projected horizontally from the top of a cliff 45\text{ m} high. If it lands on … — JAMB UTME

A body is projected horizontally from the top of a cliff $45\text{ m}$ high. If it lands on flat ground at a horizontal distance of $120\text{ m}$ from the base of the cliff, what is the speed of the body just before it strikes the ground? (Take $g = 10\text{ m/s}^2$ )

Answer: 50 m/s

Answer

The speed of the body just before striking the ground is 50 m/s.

The time of fall is determined by the height of

45\text{ m}

, yielding

t = \sqrt{2h/g} = 3\text{ s}

. The horizontal speed is constant at

120 / 3 = 40\text{ m/s}

. The vertical velocity gained on impact is

v_y = gt = 30\text{ m/s}

. Combining these mutually perpendicular velocity components gives a final impact speed of

v = \sqrt{40^2 + 30^2} = 50\text{ m/s}

Step-by-Step Solution

Calculate the time of flight from the vertical height

t = 3 s

Vertical acceleration is constant under gravity while initial vertical velocity is zero.

Compute the constant horizontal component of velocity

v_x = 40 m/s

Horizontal speed is uniform because zero horizontal force acts on the projectile.

Compute the final vertical component of velocity at impact

v_y = 30 m/s

Vertical speed increases linearly with time due to gravitational acceleration.

Determine the magnitude of the resultant velocity vector

v = 50 m/s

The horizontal and vertical components are perpendicular, so their vector sum uses the Pythagorean theorem.

Key Concept

Horizontal Projection and Impact Velocity Vector

Estimated Time:1m 30s

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