A toy car is pushed the edge of a table with an initial horizontal velocity of 2 m/s. If the table is 1 meter high, how far horizontally from the edge of the table does the toy car land?

Ignore resistive forces. What is the final velocity of the car just before it hits the floor?

1 Answer
Morgan
Aug 20, 2017

The toy car will land #~~0.9"m"# from the table.

Explanation:

This problem can be solved using kinematics.

We have the following information:

  • #|->v_(ix)=2"m"//"s"#
  • #|->y_i=1"m"#
  • #|->y_f=0"m"#
  • #|->g=9.81"m"//"s"^2#

#=>#We can calculate the time #Deltat# it took for the toy car to hit the floor using the perpendicular (y, vertical) components, and then use that value along with the known parallel (x, horizontal) component to calculate the range #Deltax#

#color(blue)(y_f=y_i+v_(iy)Deltat+1/2a_y(Deltat)^2)#

  • Initially, there was no vertical component of the car's motion, as it moved across a horizontal surface. Therefore, #v_(iy)=0#.

  • When the car falls, we are to assume that only the force of gravity is acting upon it. Therefore, #F_"net"=F_g#, and the acceleration is #a=-g=-9.8" m"//"s"^2#.

#=>cancel(y_f)=y_i+cancel(v_(iy)Deltat)+1/2a_y(Deltat)^2#

#=>0=y_i+1/2a_y(Deltat)^2#

Solving for #Deltat#:

#color(blue)(Deltat=sqrt((-2y_i)/a_y))#

Using our known values:

#Deltat=sqrt((-2*1"m")/(-9.81" m"//"s"^2))#

#=>color(blue)(Deltat=0.452"s")#

  • We can now use a kinematic equation to calculate #Deltax#.

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

  • It is important to realize that after the car falls, it has no horizontal acceleration. Therefore, #a=0#.

#=>Deltax=v_(ix)Deltat#

Using our known values:

#Deltax=(2"m"//"s")(0.452"s")#

#=>Deltax=0.904"s"#

#=>color(blue)(Deltax~~0.9"m"#

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