Question #700d5

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Question #700d5

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Answer 1 · 2016-02-24T01:56:19

$v_{x} = 30 m s^{- 1}$ $v_x=30ms^-1$ .
$v_{y} = - 14 m s^{- 1}$ $v_y=-14ms^-1$

Explanation:

We observe that the projectile when it just hits the ground will have velocity $v_{x}$ $v_x$ along the $x$ $x$ -axis, the velocity with which it was launched.
$v_{x} = 30 m s^{- 1}$ $v_x=30ms^-1$ .

Also it will have velocity $v_{y}$ $v_y$ along the $y$ $y$ -axis due to free fall downwards under gravity.

Assuming that air resistance is negligible. And noticing that both velocities are perpendicular to each other and therefore, independent of one another, there is no change in the velocity $v_{x}$ $v_x$ along the $x$ $x$ -axis.
To calculate velocity $v_{y}$ $v_y$ along the $y$ $y$ -axis.
The projectile has a potential energy when it is launched which is given by

$P E = m g h$ $PE=mgh$

As the ball falls down, due to the Law of Conversion of Energy, this gets converted in to the kinetic energy of the projectile in the $y$ $y$ direction. (The kinetic energy of the projectile in the $x$ $x$ direction has been considered separately above.) This is given by
$K E = \frac{1}{2} m v_{y}^{2}$ $KE=1/2mv_y^2$

Equating the two at a point when projectile just hits ground $(h = 0),$ $(h=0),$ all the potential energy has been changed into kinetic energy.
$\frac{1}{2} m v_{y}^{2} = m g h$ $1/2mv_y^2=mgh$
or $v_{y} = - \sqrt{2 g h}$ $v_y=-sqrt(2gh)$ ,
$- v e$ $-ve$ sign in front of velocity indicates that it is along the $- y$ $-y$ axis.

We know that acceleration due to gravity $g = 9.81 m s^{- 2}$ $g=9.81ms^-2$ and given is $h = 10 m$ $h=10m$
We obtain
$v_{y} = - \sqrt{2 \times 9.8 \times 10}$ $v_y=-sqrt(2xx9.8xx10)$
$v_{y} = - 14 m s^{- 1}$ $v_y=-14ms^-1$

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Question #700d5

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