An object is launched vertically upward from the ground with an initial velocity of 40\tex… — JAMB UTME

An object is launched vertically upward from the ground with an initial velocity of $40\text{ m/s}$ . At the exact same instant, a second object is dropped from rest from a height of $100\text{ m}$ directly above the first object. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , at what time (in seconds) after launch will the two objects meet?

Answer: 2.5 s

Answer

The two objects meet after 2.5 seconds.

Because both objects experience identical downward gravitational acceleration (

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

), their relative acceleration is zero. The relative velocity between them remains constant at

40\text{ m/s}

. The time to cover the initial separation distance of

100\text{ m}

is calculated directly as

t = \frac{\text{distance}}{\text{relative velocity}} = \frac{100}{40} = 2.5\text{ s}

Step-by-Step Solution

Set up the position-time equations for both objects taking ground level as

y = 0\text{ m}

For the upward-launched object:

y_1(t) = u t - \frac{1}{2}gt^2 = 40t - 5t^2

. For the dropped object:

y_2(t) = h_0 - \frac{1}{2}gt^2 = 100 - 5t^2

Kinematic equations of motion under uniform gravitational acceleration apply to both bodies.

Equate the two vertical position equations to solve for the meeting time

t

40t - 5t^2 = 100 - 5t^2 \implies 40t = 100

When the objects meet, they share the exact same vertical position

y_1(t) = y_2(t)

at time

t

Calculate the value of

t

t = \frac{100}{40} = 2.5\text{ s}

Direct algebraic division yields the time elapsed before collision/meeting.

Key Concept

Relative vertical motion under uniform gravity

Estimated Time:1m 30s

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