# Question dc83d

Jun 1, 2016

${E}_{p} / {E}_{k} = {\sec}^{2} \alpha - 1$

#### Explanation:

v=v_x=v_i*cos alpha" (Horizontal component of velocity of object)"#

${E}_{k} = \frac{1}{2} \cdot m \cdot {v}_{x}^{2}$

${E}_{k} = \frac{1}{2} \cdot m \cdot {v}_{i}^{2} \cdot {\cos}^{2} \alpha$

${h}_{m} = {v}_{i}^{2} \cdot {\sin}^{2} \frac{\alpha}{2 \cdot g} \text{ maximum height of object}$

${E}_{p} = m \cdot g \cdot {h}_{m}$

${E}_{p} = m \cdot \cancel{g} \cdot \frac{{v}_{i}^{2} \cdot {\sin}^{2} \alpha}{2 \cdot \cancel{g}}$

${E}_{p} = \frac{m {v}_{i}^{2} \cdot {\sin}^{2} \alpha}{2}$

${E}_{p} / {E}_{k} = \frac{\cancel{\frac{1}{2} \cdot m} \cdot \cancel{{v}_{i}^{2}} \cdot {\sin}^{2} \alpha}{\cancel{\frac{1}{2} \cdot m} {\cancel{{v}_{i}}}^{2} \cdot {\cos}^{2} \alpha}$

${E}_{p} / {E}_{k} = {\sin}^{2} \frac{\alpha}{\cos} ^ 2 \alpha$

${\sin}^{2} \alpha = 1 - {\cos}^{2} \alpha$

${E}_{p} / {E}_{k} = \frac{1 - {\cos}^{2} \alpha}{{\cos}^{2} \alpha}$

${E}_{p} / {E}_{k} = \left(\frac{1}{\cos} ^ 2 \alpha\right) - 1$

$\frac{1}{\cos} ^ 2 \alpha = {\sec}^{2} \alpha$

${E}_{p} / {E}_{k} = {\sec}^{2} \alpha - 1$

Jan 21, 2017

As referred to the figure above.

Maximum height of the object having mass $m$ will be reached when $y$- or $\sin \theta$ component of projectile's velocity becomes $0$ (zero).

As such kinetic energy at this point is given by
$K {E}_{\text{max height}} = \frac{1}{2} m {\left(u \cos \theta\right)}^{2}$

At this point all the kinetic energy due to $y$-component of initial velocity has got converted into its Potential Energy which is also given as $m g h$
$P {E}_{\text{max height}} = m g h = \frac{1}{2} m {\left(u \sin \theta\right)}^{2}$

Ratio of the two is $\left(P {E}_{\text{max height")/(KE_"max height}}\right) = \frac{\frac{1}{2} m {u}^{2} {\sin}^{2} \theta}{\frac{1}{2} m {u}^{2} {\cos}^{2} \theta}$
$= {\tan}^{2} \theta$