# Question #8efea

Aug 3, 2016

The ball is $\frac{3 h}{4}$ meters from the ground at $t = \frac{T}{2}$

#### Explanation:

I'm neglecting air resistance, but can add it in upon request. The body moves in only one dimension, which we shall call $y$. We shall define the origin of our coordinate system at the ground ($y = 0$) and the initial position of the body $y \left(0\right) = h$. As is the usual convention we define upwards as positive.

Without air resistance, the only force acting on the body will be gravity - this acts in the negative y direction to provide an unbalanced force which we can use Newton's second law on:

$m \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = - m g$

$\frac{{d}^{2} y}{{\mathrm{dt}}^{2}}$ represents acceleration in the y direction.

If you are unfamiliar with using calculus to solve problems like this then skip to the blue heading, this is just the derivation of a formula common to introductory dynamics.

$\therefore \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = - g$

Integrating wrt t gives:

$\frac{\mathrm{dy}}{\mathrm{dt}} = - g t + C$

The body starts from rest so $y ' \left(0\right) = 0$.

$\implies C = 0$

$\frac{\mathrm{dy}}{\mathrm{dt}} = - g t$

Integrate again wrt t:

$y \left(t\right) = - \frac{1}{2} g {t}^{2} + C$

$y \left(0\right) = h = C$

$\implies y \left(t\right) = h - \frac{1}{2} g {t}^{2}$

This is our expression for the displacement from the origin as a function of time. Notice it's very similar to the common expression

$s = {v}_{0} t + \frac{1}{2} a {t}^{2}$

which describes displacement under constant acceleration. This is because it is the exact same derivation, we have just accounted for the fact that initial speed is zero while doing it.

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Anyway, onto solving the problem:

$y \left(t\right) = h - \frac{1}{2} g {t}^{2}$

If it takes T seconds to reach the ground, $y \left(T\right) = 0$.

$y \left(T\right) = 0 = h - \frac{1}{2} g {T}^{2}$

$\implies h = \frac{1}{2} g {T}^{2}$

${T}^{2} = \frac{2 h}{g}$

$T = \sqrt{\frac{2 h}{g}}$

Note that we discard the negative solution from the square root because negative time doesn't really make sense.

$\frac{T}{2} = \frac{1}{2} \sqrt{\frac{2 h}{g}} = \sqrt{\frac{h}{2 g}}$

$y \left(\frac{T}{2}\right) = h - \frac{1}{2} g {\left(\sqrt{\frac{h}{2 g}}\right)}^{2}$

$y \left(\frac{T}{2}\right) = h - \frac{1}{2} g \left(\frac{h}{2 g}\right) = h - \frac{h}{4} = \frac{3 h}{4}$

Aug 4, 2016

The body is at $\frac{h}{4} m$ from the top of tower.

#### Explanation:

Kinematic equation for the body dropped from a height $h$ is
$h = u t + \frac{1}{2} g {t}^{2}$
where $u , g \mathmr{and} t$ are initial velocity, acceleration due to gravity and time taken to reach ground respectively.
Inserting given values we get
$h = 0 \times T + \frac{1}{2} g {T}^{2}$
$\implies h = \frac{1}{2} g {T}^{2}$ .......(1)

Similarly height ${h}_{1}$ dropped in time $\frac{T}{2}$ is
${h}_{1} = \frac{1}{2} g {\left(\frac{T}{2}\right)}^{2}$
$\implies {h}_{1} = \frac{1}{8} g {T}^{2}$......(2)
Dividing (2) by (1) we get
${h}_{1} / h = \frac{1}{4}$

$\implies {h}_{1} = \frac{h}{4}$