A projectile is launched from ground level over flat terrain with a constant horizontal ve… — JAMB UTME

A projectile is launched from ground level over flat terrain with a constant horizontal velocity component of $10\text{ m/s}$ . At a height of $40\text{ m}$ above the ground, the magnitude of its vertical velocity component is equal to its horizontal velocity component. Taking $g = 10\text{ m/s}^2$ , calculate the maximum height reached by the projectile in meters.

Answer: 45 m

Answer

The maximum height reached by the projectile is

45\text{ m}

At the height of

40\text{ m}

, the vertical velocity is equal to the horizontal velocity of

10\text{ m/s}

. Applying the vertical kinematic relation

v_y^2 = u_y^2 - 2gh

yields

10^2 = u_y^2 - 2(10)(40)

, which gives

u_y^2 = 900\text{ m}^2/\text{s}^2

. The maximum height attained above ground level occurs when

v_y = 0

, calculated as

H_{\text{max}} = \frac{u_y^2}{2g} = \frac{900}{20} = 45\text{ m}

Step-by-Step Solution

Identify the vertical component of velocity at the given height

v_y = 10\text{ m/s}

h = 40\text{ m}

The problem states that at

h = 40\text{ m}

, the vertical velocity component equals the constant horizontal component

u_x = 10\text{ m/s}

Determine the initial vertical launch velocity component

u_y

u_y^2 = 900\text{ m}^2/\text{s}^2 \implies u_y = 30\text{ m/s}

Applying the vertical motion equation

v_y^2 = u_y^2 - 2gh

gives

10^2 = u_y^2 - 2(10)(40) \implies u_y^2 = 100 + 800 = 900

Calculate the maximum height

H_{\text{max}}

H_{\text{max}} = 45\text{ m}

At maximum height, the vertical velocity becomes zero. Using

H_{\text{max}} = \frac{u_y^2}{2g} = \frac{900}{20} = 45\text{ m}

Key Concept

Vertical Kinematics and Maximum Height of a Projectile

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