An object is projected from level ground with an initial speed of 30\text{ m/s} at an angl… — JAMB UTME

An object is projected from level ground with an initial speed of $30\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the maximum height reached by the object in meters.

Answer: 11.25 m

Answer

The maximum height reached by the object is

11.25\text{ m}

The vertical component of initial velocity is

u_y = 30 \sin(30^\circ) = 15\text{ m/s}

. At maximum height, vertical velocity becomes zero, so

H = \frac{u_y^2}{2g} = \frac{15^2}{20} = 11.25\text{ m}

Step-by-Step Solution

Calculate the vertical component of the initial velocity.

u_y = u \sin\theta = 30 \times \sin(30^\circ) = 15\text{ m/s}

Only the vertical component of initial velocity determines the maximum height.

Apply the vertical motion equation at maximum height where vertical velocity is zero.

H = \frac{u_y^2}{2g} = \frac{15^2}{2 \times 10} = 11.25\text{ m}

Using

v_y^2 = u_y^2 - 2gH

with

v_y = 0

gives

H = \frac{u_y^2}{2g}

Key Concept

Maximum height of a projectile

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