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 ball is projected in the air with initial speed

u

and goes up and down with acceleration

g

downwards. A timer is at a height

h

. It records the time from the ball being projected until it passes the timer on the way up as

t_{1}

and on the way down as

t_{2}

. Show that the total of the two times is independent of

h

and that the initial speed can be calculated as

u = \frac{g (t_{1} + t_{2})}{2}

. Show also that the difference between the times is given by

\frac{2 \sqrt{u^{2} - 2 g h}}{g}

. Hence, find a formula for

h

in terms of

t_{1,} t_{2}

and

g

.

Accepted Answer

Here, use the constant acceleration formula for the gravity as follows

$h = u t - \frac{1}{2} g t^{2}$

$u = \frac{g (t_{1} + t_{2})}{2}$

$\frac{2 \sqrt{u^{2} - 2 g h}}{g}$

$t_{1} = \frac{2 u - \sqrt{4 u^{2} - 8 g h}}{2 g}$

$t_{2} = \frac{2 u + \sqrt{4 u^{2} - 8 g h}}{2 g}$

$t_{1} + t_{2} = \frac{2 u - \sqrt{4 u^{2} - 8 g h} + 2 u + \sqrt{4 u^{2} - 8 g h}}{2 g}$

$= \frac{4 u}{2 g} = \frac{2 u}{g}$

$u = \frac{g (t_{1} + t_{2})}{2}$

$t_{2} - t_{1} = \frac{- (2 u - \sqrt{4 u^{2} - 8 g h}) + 2 u + \sqrt{4 u^{2} - 8 g h}}{2 g}$

$= \frac{2 \sqrt{4 u^{2} - 8 g h}}{2 g}$

$= \frac{4 \sqrt{u^{2} - 2 g h}}{2 g}$

$= \frac{2 \sqrt{u^{2} - 2 g h}}{g}$

Note that near the end of this calculation we pull out a factor of root 4. when taking the square root of an expression, taking out any square factors will help you to simplify things in this way.

We now have

$t_{2} - t_{1} = \frac{2 \sqrt{u^{2} - 2 g h}}{g}$

$g (t_{2} - t_{1}) = 2 \sqrt{u^{2} - 2 g h}$

$\sqrt{u^{2} - 2 g h} = \frac{g (t_{2} - t_{1})}{2}$

$u^{2} - 2 g h = \frac{g^{2} {(t_{2} - t_{1})}^{2}}{4}$

But

$u^{2} = \frac{g^{2} {(t_{1} + t_{2})}^{2}}{4}$

$\frac{g^{2} {(t_{1} + t_{2})}^{2}}{4} - 2 g h = \frac{g^{2} {(t_{2} - t_{1})}^{2}}{4}$

$8 g h = g^{2} {(t_{1} + t_{2})}^{2} - g^{2} {(t_{2} - t_{1})}^{2}$

$8 g h = g^{2} ({t_{1}}^{2} + {t_{2}}^{2} - 2 t_{1} t_{2} - {t_{1}}^{2} - {t_{2}}^{2} + 2 t_{1} t_{2})$

$8 g h = 4 g^{2} t_{1} t_{2}$

$h = \frac{4 g^{2} t_{1} t_{2}}{8 g} = \frac{g t_{1} t_{2}}{2}$

Important Questions on Velocity and Acceleration

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