Name: Solving Free Fall Problem Using Quadratic Equation
Uploaded: 2019-07-19
Duration: 7 min 7 s
Description: In this video, we will learn how we can solve the problem of free fall effectively using the quadratic equation. Let's watch the video and learn.

Question

A footballer kicks a ball directly towards a wall

10 m

away and walks after the ball in the same direction at a constant

2 m s^{- 1} .

The ball starts at

4 m s^{- 1}

but decelerates at a constant rate of

0.5 m s^{- 2} .

When it hits the wall it rebounds to travel away from the wall at the same speed with which it hit the wall.
Find the time after the initial kick when the ball returns to the footballer.

Accepted Answer

In this question, there is a sudden change in the direction of the ball which means acceleration is not constant.

Time and speed at which ball hits the wall,

$2 m s^{- 1} .$

First time, $t = 8 - 2 \sqrt{6} s$

$v = u + a t \Rightarrow v = 4 - 0.5 (8 - 2 \sqrt{6}) \Rightarrow v = 4 - 4 + \sqrt{6} \Rightarrow v = \sqrt{6} {ms}^{- 1}$

During this motion, the footballer walks towards the wall,

$s = v t \Rightarrow s = 2 (8 - 2 \sqrt{6}) \Rightarrow s = 16 - 4 \sqrt{6} m$

The distance between the ball and the footballer is,

$\Rightarrow 10 - (16 - 4 \sqrt{6}) = 4 \sqrt{6} - 6 m$

Now, If the time taken for the ball to return to the footballer after rebounding is $t_{2}$ seconds,

Distance moved by the footballer $+$ Distance moved by the ball $= 4 \sqrt{6} - 6$

$\Rightarrow 2 t_{2} + \sqrt{6} t_{2} - \frac{1}{2} \times 0.5 t_{2}^{2} = 4 \sqrt{6} - 6 \Rightarrow (2 + \sqrt{6}) t_{2} - 0.25 t_{2}^{2} = 4 \sqrt{6} - 6 \Rightarrow t_{2}^{2} - 4 (2 + \sqrt{6}) t_{2} + 16 \sqrt{6} - 24 = 0$

On solving this,

$t_{2} = 0.8989 s or 16.898 s$

First returns to the footballer when, $t_{2} = 0.8989 s$

So, total time will be $t_{2} + 8 - 2 \sqrt{6} = 4.00 s$

Important Questions on Velocity and Acceleration

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