Question

A boy, standing at a certain height, kicks a football horizontally with a velocity of $16\text{ m s}^{-1}$. What will be the ratio of horizontal and vertical components of velocities after $2\text{ s}$? (Take $g = 8\text{ m s}^{-2}$)}

• A. $1:1$
• B. $(\frac{\sqrt{3}}{2}):1$
• C. $1:\sqrt{2}$
• D. $1:0.5$

Hint:

Whenever the horizontal velocity matches the value of $(g \times t)$, the components will always be equal, meaning the path creates a $45^\circ$ angle relative to the horizon at that specific point in time!

Answer 1

The Correct Option is A

Concept: This problem models horizontal projectile motion. In the absence of air resistance, there is no force acting horizontally, so the horizontal acceleration is $0$ ($a_x = 0$). This ensures the horizontal velocity component remains constant throughout the flight time. Vertically, the object is subject to a constant gravitational acceleration downward ($a_y = g$).

Step 1: Evaluate the horizontal component of velocity ($v_x$).
The object is kicked with an initial horizontal velocity $u_x = 19.6\text{ m s}^{-1}$. Since $a_x = 0$: \[ v_x = u_x = 19.6\text{ m s}^{-1} \]

Step 2: Evaluate the vertical component of velocity ($v_y$) at $t = 2\text{ s}$.
Initially, the projectile has no vertical movement, so $u_y = 0$. Using the first equation of motion along the vertical axis: \[ v_y = u_y + gt \] Substitute $u_y = 0$, $g = 9.8\text{ m s}^{-2}$, and $t = 2\text{ s}$: \[ v_y = 0 + (9.8 \times 2) = 19.6\text{ m s}^{-1} \]

Step 3: Compute the ratio of horizontal to vertical components.
The desired ratio is: \[ \text{Ratio} = \frac{v_x}{v_y} = \frac{19.6}{19.6} = \frac{1}{1} \implies 1:1 \]