A stone is projected vertically upwards from the top edge of a cliff 80\text{ m} high with… — JAMB UTME

A stone is projected vertically upwards from the top edge of a cliff $80\text{ m}$ high with an initial speed of $30\text{ m/s}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the total time, in seconds, taken by the stone to reach the ground at the base of the cliff.

Answer: 8 s

Answer

The total time taken by the stone to reach the ground at the base of the cliff is

8\text{ s}

Using the equation of motion

s = ut - \frac{1}{2}gt^2

with

s = -80\text{ m}

u = 30\text{ m/s}

, and

g = 10\text{ m/s}^2

yields the quadratic equation

t^2 - 6t - 16 = 0

. Solving gives

t = 8\text{ s}

(ignoring the unphysical negative root

t = -2\text{ s}

). Alternatively, breaking the motion into two parts: time to reach maximum height (

30\text{ m/s} / 10\text{ m/s}^2 = 3\text{ s}

, covering

45\text{ m}

) plus time to fall from maximum height of

125\text{ m}

to the ground (

t = \sqrt{2(125)/10} = 5\text{ s}

), giving a total time of

3 + 5 = 8\text{ s}

Step-by-Step Solution

Set up the kinematic equation with appropriate vector signs

Displacement

s = -80\text{ m}

, initial velocity

u = +30\text{ m/s}

, acceleration

a = -g = -10\text{ m/s}^2

Since the ground is below the release point, displacement is negative when taking the upward direction as positive.

Substitute values into

s = ut + \frac{1}{2}at^2

-80 = 30t - 5t^2

Relates displacement, initial speed, time, and constant gravitational acceleration.

Form and solve the quadratic equation

5t^2 - 30t - 80 = 0 \implies t^2 - 6t - 16 = 0 \implies (t - 8)(t + 2) = 0

Simplifies the algebraic expression to find the time roots.

Select the physical root

t = 8\text{ s}

Time elapsed must be a positive quantity.

Key Concept

Kinematics of Vertical Motion under Gravity with Displacement from Elevation

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic