A stone is thrown vertically downwards from the top of a 60\text{ m} high tower with an in… — JAMB UTME

A stone is thrown vertically downwards from the top of a $60\text{ m}$ high tower with an initial speed of $5\text{ m/s}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the time taken, in seconds, for the stone to reach the ground.

Answer: 3 s

Answer

The stone takes

3\text{ s}

to reach the ground.

Applying the equation

h = ut + \frac{1}{2}gt^2

with

h = 60\text{ m}

u = 5\text{ m/s}

, and

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

yields

60 = 5t + 5t^2

. Dividing by

5

gives

t^2 + t - 12 = 0

, which factorizes into

(t + 4)(t - 3) = 0

. Rejecting

t = -4\text{ s}

leaves the correct time of

3\text{ s}

Step-by-Step Solution

Set up the vertical motion equation

Using

h = ut + \frac{1}{2}gt^2

, substitute

h = 60\text{ m}

u = 5\text{ m/s}

, and

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

This equation directly relates displacement, initial speed, acceleration, and time.

Form and solve the quadratic equation for time

60 = 5t + 5t^2 \Rightarrow t^2 + t - 12 = 0 \Rightarrow (t - 3)(t + 4) = 0

, giving

t = 3\text{ s}

Time must be positive, so the physically meaningful solution is

t = 3\text{ s}

Key Concept

Vertical motion under gravity with non-zero initial downward velocity

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