A projectile is launched from ground level with an initial speed of 40\text{ m/s} at an an… — JAMB UTME

A projectile is launched from ground level with an initial speed of $40\text{ m/s}$ at an angle $\theta$ to the horizontal such that $\sin\theta = 0.8$ and $\cos\theta = 0.6$ . Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the speed of the projectile in $\text{m/s}$ when it reaches a height of $35\text{ m}$ above the ground?

Answer: 30 m/s

Answer

The speed of the projectile at a height of

35\text{ m}

above the ground is

30\text{ m/s}

The correct answer is

30\text{ m/s}

. The vertical component of velocity at height

35\text{ m}

is found using

v_y^2 = u_y^2 - 2gh = 32^2 - 2(10)(35) = 324

, giving

v_y = 18\text{ m/s}

. Since horizontal velocity component remains constant at

v_x = 24\text{ m/s}

, the overall speed magnitude is given by

v = \sqrt{v_x^2 + v_y^2} = \sqrt{24^2 + 18^2} = \sqrt{900} = 30\text{ m/s}

Step-by-Step Solution

Resolve initial velocity into horizontal and vertical components

u_x = 24\text{ m/s}

and

u_y = 32\text{ m/s}

Horizontal component

u_x = u\cos\theta = 40 \times 0.6 = 24\text{ m/s}

and vertical component

u_y = u\sin\theta = 40 \times 0.8 = 32\text{ m/s}

Calculate vertical velocity component at height

h = 35\text{ m}

v_y = 18\text{ m/s}

Using

v_y^2 = u_y^2 - 2gh

, we get

v_y^2 = 32^2 - 2(10)(35) = 1024 - 700 = 324

, yielding

v_y = 18\text{ m/s}

Calculate the magnitude of total velocity at height

h = 35\text{ m}

v = 30\text{ m/s}

Because air resistance is neglected, horizontal velocity remains constant (

v_x = u_x = 24\text{ m/s}

). Total speed is

v = \sqrt{v_x^2 + v_y^2} = \sqrt{24^2 + 18^2} = 30\text{ m/s}

Key Concept

Independence of perpendicular velocity components and calculation of instantaneous speed in projectile motion.

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