# Question #95d72

May 7, 2016

$793 m$

#### Explanation:

The velocity of projection of the bag is $u = 90 \frac{m}{s}$

The angle of projection of the suitcase is $\alpha = {23}^{\circ}$

Initial vertical component of the velocity(upward)$= u \sin \alpha$

The vertical displacement of the bag , $h = - 114 m$
Acceleration due to gravity $g = - 9.8 \frac{m}{s} ^ 2$
" since upward initial vertical component of velocity taken +ve"

The horizontal component of the velocity$= u \cos \alpha$ will remain unaltered until the object lands

Let the time required for its landing after it is thrown is T sec

So applying equation of motion under gravity we can write

$h = u \sin \alpha \times T - \frac{1}{2} g \times {T}^{2}$
$\implies - 114 = 90 \sin 23 \times T - \frac{1}{2} \times 9.8 \times {T}^{2}$
$\implies 4.9 {T}^{2} - 35 T - 114 = 0$
$T = \frac{- \left(- 35\right) + \sqrt{{\left(- 35\right)}^{2} - 4 \times 4.9 \left(- 114\right)}}{2 \times 4.9}$
$T = 9.57 s$
Horizontal displacent of suitcase during this time of fall is $u \cos \alpha \times T = 90 \cdot \cos 23 \cdot 9.57 = 793 m$
So the suitcase will land at a distance of 793m from the dog.