A projectile is launched from ground level with a horizontal velocity component of 15\text… — JAMB UTME

A projectile is launched from ground level with a horizontal velocity component of $15\text{ m/s}$ and a vertical velocity component of $20\text{ m/s}$ . Neglecting air resistance, what is the magnitude of the velocity of the projectile at its maximum height?

$15\text{ m/s}$ Answer
B
$0\text{ m/s}$
C
$20\text{ m/s}$
D
$25\text{ m/s}$

Answer

The magnitude of the velocity of the projectile at its maximum height is

15\text{ m/s}

In projectile motion under gravity without air resistance, the horizontal component of velocity remains constant throughout flight (

v_x = 15\text{ m/s}

). At the highest point (apex), the vertical component of velocity momentarily becomes zero (

v_y = 0\text{ m/s}

). Therefore, the magnitude of the velocity at the maximum height is equal to the horizontal component, which is

15\text{ m/s}

Step-by-Step Solution

Identify the velocity components at the maximum height of a projectile

Vertical component

v_y = 0\text{ m/s}

and horizontal component

v_x = u_x = 15\text{ m/s}

Gravity acts vertically, reducing vertical velocity to zero at the peak, while horizontal velocity remains constant in the absence of air resistance.

Calculate the total magnitude of velocity at maximum height

v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 0^2} = 15\text{ m/s}

The magnitude of the resultant velocity vector is derived using the Pythagorean theorem.

Key Concept

Velocity at Maximum Height in Projectile Motion

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