Question

Question #95d72

Answer 1

`793m`

Explanation:

The velocity of projection of the bag is `u = 90m/s`

The angle of projection of the suitcase is `alpha = 23^@`

Initial vertical component of the velocity(upward)`=usinalpha`

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

The horizontal component of the velocity`=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=usinalphaxxT-1/2gxxT^2`
`=>-114=90sin23xxT-1/2xx9.8xxT^2`
`=>4.9T^2-35T-114=0`
`T =( -(-35)+sqrt((-35)^2-4xx4.9(-114)))/(2xx4.9)`
`T=9.57s`
Horizontal displacent of suitcase during this time of fall is `ucosalphaxxT=90*cos23*9.57=793m`
So the suitcase will land at a distance of 793m from the dog.