Question

Q1.Prove mathematically when a ball is projected upward then it velocity decreases with increasing height. ​

Answer 1

Explanation:

Let us consider the ball to be projected upward with a velocity u.

The force of gravity decelerates the ball, and finally at Maximum height, Velocity becomes zero.

For this situation,

Initial velocity = u

Acceleration = - g

[Considering the upward direction as positive]

Let the velocity of the object in its upward motion at a height x above the ground is v

From Third equation of motion,

v² - u² = 2(-g)(x)

⇒ v² = u² - 2gx

⇒ v = √(u²-2gx)

y(x) decreases with x if, y'(x) < 0

v(x) = √(u² - 2gx)

Differentiating with respect to x on both sides,

$\begin{gathered} \frac{dv}{dx} = \frac{1}{2 \sqrt{ {u}^{2} - 2gx} } \times - 2g \\ \\ \frac{dv}{dx} = \frac{ - g}{v} \end{gathered} </p><p>$

We are considering only the upward motion of the ball, so v > 0

g being a constant is always positive.

So, v'(x) < 0

Thus, Mathematically we proved that, For a ball projected upward, It's velocity decreases with increasing height.