Question #95d72

Question

Question #95d72

1 Answer

Impact of this question

3626 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-07T14:52:08

$793 m$ $793m$

Explanation:

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

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

Initial vertical component of the velocity(upward) $= u sin α$ $=usinalpha$

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

The horizontal component of the velocity $= u cos α$ $=ucosalpha$ 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

$h = u sin α \times T - \frac{1}{2} g \times T^{2}$ $h=usinalphaxxT-1/2gxxT^2$
$\Rightarrow - 114 = 90 sin 23 \times T - \frac{1}{2} \times 9.8 \times T^{2}$ $=>-114=90sin23xxT-1/2xx9.8xxT^2$
$\Rightarrow 4.9 T^{2} - 35 T - 114 = 0$ $=>4.9T^2-35T-114=0$
$T = \frac{- (- 35) + \sqrt{{(- 35)}^{2} - 4 \times 4.9 (- 114)}}{2 \times 4.9}$ $T =( -(-35)+sqrt((-35)^2-4xx4.9(-114)))/(2xx4.9)$
$T = 9.57 s$ $T=9.57s$
Horizontal displacent of suitcase during this time of fall is $u cos α \times T = 90 \cdot cos 23 \cdot 9.57 = 793 m$ $ucosalphaxxT=90*cos23*9.57=793m$
So the suitcase will land at a distance of 793m from the dog.

Science

Math

Humanities

... and beyond

Question #95d72

1 Answer

Explanation:

Related questions

Impact of this question