A stone is thrown into the air with an initial velocity of 20\text{ m/s} at an angle of 60… — JAMB UTME

A stone is thrown into the air with an initial velocity of $20\text{ m/s}$ at an angle of $60^\circ$ to the horizontal. What is the magnitude of its velocity at the highest point of its trajectory?

A
$0\text{ m/s}$
$10\text{ m/s}$ Answer
C
$17.3\text{ m/s}$
D
$20\text{ m/s}$

Answer

The magnitude of the velocity at the highest point is

10\text{ m/s}

At the peak of a projectile's trajectory, the vertical velocity component vanishes (

v_y = 0\text{ m/s}

), but horizontal motion continues at a constant speed (

v_x = u \cos \theta

). Substituting

u = 20\text{ m/s}

and

\theta = 60^\circ

yields

v_x = 20 \times 0.5 = 10\text{ m/s}

Step-by-Step Solution

Identify the velocity components at the apex

At maximum height, the vertical velocity component is

v_y = 0\text{ m/s}

, while the horizontal component remains constant at

v_x = u_x = u \cos \theta

Horizontal acceleration is zero when air resistance is neglected.

Calculate the horizontal component of velocity

v_x = 20 \times \cos(60^\circ) = 20 \times 0.5 = 10\text{ m/s}

The trigonometric cosine function gives the horizontal projection of the initial velocity vector.

Determine total velocity magnitude at the apex

v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 0^2} = 10\text{ m/s}

Since the vertical component is zero, the total velocity at the peak equals the horizontal component.

Key Concept

Velocity components at maximum height in projectile motion

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