Question #e6fe4

1 Answer
Morgan
Aug 17, 2017

Should stand around #7.2"m"# from the base of the building.

Explanation:

This can be solved using kinematics.

We have the following information:

  • #|-> v=4.5"m"//"s"#
  • #|->theta=25^o#
  • #|->h_i=12"m"#
  • #|->h_f~~0"m"#
  • #|->a=-g=-9.8"m"//"s"^2#

We will have to decompose the velocity vector into its parallel (x, horizontal) and perpendicular (y, vertical) components, and use multiple kinematic equations to solve for the displacement.

Diagram:

enter image source here

We can use trigonometry to find the components of the initial velocity vector.

#sin(theta)="opposite"/"hypotenuse"#

#=>sin(theta)=v_y/v#

#=>v_y=vsin(theta)#

#=>v_y=4.5sin(25^o)#

#=>color(darkblue)(v_y=1.902"m"//"s")#

Similarly, we find that #color(darkblue)(v_x=4.078"m"//"s")#

We can use the perpendicular component of velocity to calculate the flight time of the beanbag, knowing that for an object in free fall, #a=-g#.

#h_f-h_i=v_(iy)Deltat+1/2a(Deltat)^2#

We want to solve for #Deltat#, which can be done using the quadratic equation.

#=>0=-4.9t^2+1.90v+12#

#t=(-b+-sqrt(b^2-4ac))/(2a)#

#=>t=(-1.9+-sqrt((1.9)^2-4(-4.9)(12)))/(-9.8)#

#=>color(darkblue)(t~~1.77"s")#

Using the flight time, we can now calculate the range of the projectile using the following kinematic:

#Deltax=v_(ix)Deltat+1/2a(Deltat)^2#

We want to solve for #Deltax#.

Because there is no horizontal acceleration for an object in free fall:

#Deltax=v_(ix)Deltat#

#=>=(4.078"m"//"s")(1.77"s")#

#=>=07.221"m"#

#=>~~7.2"m"#.

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