A projectile is launched from horizontal ground with an initial speed of 60\text{ m/s} at … — JAMB UTME

A projectile is launched from horizontal ground with an initial speed of $60\text{ m/s}$ at an angle of $60^\circ$ to the horizontal. Taking $g = 10\text{ m/s}^2$ and neglecting air resistance, what is the speed of the projectile at its maximum height?

A
$0\text{ m/s}$
$30\text{ m/s}$ Answer
C
$30\sqrt{3}\text{ m/s}$
D
$60\text{ m/s}$

Answer

The speed of the projectile at its maximum height is

30\text{ m/s}

In two-dimensional projectile motion, gravity acts purely in the vertical direction. At the apex (maximum height), the vertical velocity component

v_y

drops to zero. However, the horizontal velocity component

v_x = u \cos \theta

remains constant throughout the entire flight because air resistance is neglected. Consequently, the speed at maximum height equals

u \cos 60^\circ = 60 \times 0.5 = 30\text{ m/s}

Step-by-Step Solution

Resolve initial launch velocity into orthogonal horizontal and vertical components.

u_x = u \cos \theta = 60 \cos 60^\circ = 30\text{ m/s}

and

u_y = u \sin \theta = 60 \sin 60^\circ = 30\sqrt{3}\text{ m/s}

Projectile motion consists of independent horizontal and vertical motions.

Determine the velocity components at the apex (maximum height).

At the peak, vertical velocity

v_y = 0\text{ m/s}

while horizontal velocity remains

v_x = u_x = 30\text{ m/s}

Gravity acts vertically causing

v_y

to become zero at peak height, whereas no horizontal force acts, leaving

v_x

unchanged.

Calculate the magnitude of total velocity (speed) at maximum height.

v = \sqrt{v_x^2 + v_y^2} = \sqrt{30^2 + 0^2} = 30\text{ m/s}

Speed is the resultant magnitude of orthogonal velocity components.

Key Concept

Velocity at Maximum Height in Projectile Motion

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