A body is projected vertically upwards from the ground with an initial velocity u. It pass… — JAMB UTME

A body is projected vertically upwards from the ground with an initial velocity $u$ . It passes a point at a height of $40\text{ m}$ above the ground at $t = 2\text{ s}$ while ascending and again at $t = 4\text{ s}$ while descending. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the initial speed of projection $u$ of the body?

A
$20\text{ m/s}$
$30\text{ m/s}$ Answer
C
$40\text{ m/s}$
D
$60\text{ m/s}$

Answer

The initial speed of projection of the body is

30\text{ m/s}

The position of a body thrown vertically upwards is given by

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

. Rearranging this equation into standard quadratic form gives

t^2 - \left(\frac{2u}{g}\right)t + \frac{2h}{g} = 0

. The roots

t_1

and

t_2

correspond to the times the body reaches height

h

. By Vieta's formulas, the sum of the times is

t_1 + t_2 = \frac{2u}{g}

. Substituting

t_1 = 2\text{ s}

t_2 = 4\text{ s}

, and

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

gives

6 = \frac{2u}{10}

, which solves to

u = 30\text{ m/s}

. Alternatively, the time to reach maximum height is the midpoint

t_{\text{top}} = \frac{t_1 + t_2}{2} = 3\text{ s}

, and at maximum height

v = 0 = u - gt_{\text{top}}

, yielding

u = 10 \times 3 = 30\text{ m/s}

Step-by-Step Solution

Set up the vertical motion displacement equation

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

The equation describes the vertical position

h

at any time

t

for a projectile launched from ground level with initial speed

u

Rearrange the equation into standard quadratic form for

t

\frac{1}{2}gt^2 - ut + h = 0 \implies t^2 - \left(\frac{2u}{g}\right)t + \frac{2h}{g} = 0

The two solutions

t_1

and

t_2

represent the times at which the body reaches the specific height

h

Apply Vieta's formulas for the sum of roots of the quadratic equation

t_1 + t_2 = \frac{2u}{g}

The sum of the roots of a quadratic equation

t^2 - Bt + C = 0

is equal to the coefficient

B

Substitute given values to solve for

u

2 + 4 = \frac{2u}{10} \implies 6 = \frac{u}{5} \implies u = 30\text{ m/s}

Given

t_1 = 2\text{ s}

t_2 = 4\text{ s}

, and

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

, direct substitution yields the initial velocity.

Key Concept

Vertical Motion under Gravity and Time Symmetry

Estimated Time:1m 0s

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