A projectile is launched from level ground with an initial velocity of 50\text{ m/s} at an… — JAMB UTME

A projectile is launched from level ground with an initial velocity of $50\text{ m/s}$ at an angle of $60^\circ$ to the horizontal. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the speed of the projectile at the highest point of its trajectory?

A
$0\text{ m/s}$
$25\text{ m/s}$ Answer
C
$25\sqrt{3}\text{ m/s}$
D
$50\text{ m/s}$

Answer

The speed of the projectile at its highest point is

25\text{ m/s}

In 2D projectile motion, the horizontal component of velocity remains constant throughout flight because no horizontal force acts on the object. At the highest point, vertical velocity reduces to zero, making the total speed equal exclusively to the horizontal velocity component:

v = u \cos 60^\circ = 50 \times 0.5 = 25\text{ m/s}

Step-by-Step Solution

Resolve the initial velocity into horizontal and vertical components

u_x = u \cos\theta = 50 \cos 60^\circ = 25\text{ m/s}

and

u_y = u \sin\theta = 50 \sin 60^\circ = 25\sqrt{3}\text{ m/s}

Projectile motion decomposes into independent horizontal (constant velocity) and vertical (uniform acceleration) components.

Determine the vertical component of velocity at maximum height

v_y = 0\text{ m/s}

At the apex of the parabolic path, the upward vertical motion momentarily stops before descending.

Calculate total speed at the apex using vector magnitude

v = \sqrt{v_x^2 + v_y^2} = \sqrt{25^2 + 0^2} = 25\text{ m/s}

Because horizontal acceleration is zero (ignoring air resistance),

v_x = u_x = 25\text{ m/s}

throughout the flight.

Key Concept

Horizontal Component of Velocity in Projectile Motion

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