# Question #17328

Dec 19, 2016

#### Explanation:

The object is thrown so it moves up with an initial vertical velocity of $v$ making an angle $B$ with local horizontal. Its horizontal and vertical components are given as

${v}_{x} = v \cos B$

${v}_{y} = v \sin B$

Ignoring air friction horizontal component is horizontal motion with constant velocity.

Vertical component takes the object up, reduces due to acceleration due to gravity acting in the opposite direction, becomes zero at a particular instant. Thereafter, the object falls freely.
As both the components are orthogonal these can be treated separately.

Now Vertical displacement $s = {y}_{f} - {y}_{i}$
Where ${y}_{f} - {y}_{i}$ are final $y$ coordinate and initial $y$ coordinate respectively.

When the object reaches ground we have
$s = 0 - y = - y$

Applicable kinematic equation for vertical displacement and other quantities of interest is
${v}^{2} - {u}^{2} = 2 g s$
${v}_{f y}^{2} - {\left(v \sin B\right)}^{2} = 2 \left(- g\right) \left(- y\right)$
$\implies {v}_{f y}^{2} = {\left(v \sin B\right)}^{2} + 2 g y$
$\implies {v}_{f y} = \pm \sqrt{{\left(v \sin B\right)}^{2} + 2 g y}$
Resultant velocity is sum of $x$ and $y$ components
Resultant velocity$= v \cos B \hat{x} - \sqrt{{\left(v \sin B\right)}^{2} + 2 g y} \hat{y}$

In vector notation we have selected $- v e$ sign as that is the direction of $y$ component of velocity as the object hits ground.

.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.

Alternate method of finding vertical component of velocity as the object hits ground.
At the time when object is thrown upwards, its total energy is
$K E + P E$
As it hits ground all the energy is due to its $K E$. Using Law of conservation of energy and equating both
$\frac{1}{2} m {\left(v \sin B\right)}^{2} + m g y = \frac{1}{2} m {v}_{f y}^{2}$
This equation gives you the same result
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.--.-.-.