A projectile is launched from level ground with an initial speed of 25\text{ m/s} at an an… — JAMB UTME

A projectile is launched from level ground with an initial speed of $25\text{ m/s}$ at an angle $\theta$ to the horizontal such that $\sin\theta = 0.80$ . Calculate the maximum height reached by the projectile in meters. [Take $g = 10\text{ m/s}^2$ ]

Answer: 20 m

Answer

The maximum height reached by the projectile is

20\text{ m}

The vertical component of the initial launch velocity is

u_y = u \sin\theta = 25 \times 0.80 = 20\text{ m/s}

. Using the equation for maximum height

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

, we substitute

u_y = 20\text{ m/s}

and

g = 10\text{ m/s}^2

to obtain

H = \frac{400}{20} = 20\text{ m}

Step-by-Step Solution

Calculate the initial vertical velocity component (

u_y

)

u_y = 25\text{ m/s} \times 0.80 = 20\text{ m/s}

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

Calculate the maximum height (

H

) using kinematic equations

H = \frac{u_y^2}{2g} = \frac{20^2}{2 \times 10} = 20\text{ m}

At maximum height, the vertical component of velocity becomes zero.

Key Concept

Maximum height of a projectile depends entirely on its initial vertical component of velocity and acceleration due to gravity.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic