An iron ball rolls off a cliff which is 23.3 m high and lands on your new car. Assume your car is 2.3 m tall. How fast was the iron ball traveling just before it hit your car in m/s?

1 Answer
Nathan L.
Jul 10, 2017

#v_y = 20.3# #"m/s"#

Explanation:

#2.3# #"m"# is not your average car height! Nevertheless, here's a solution:

We're asked to find the speed of the ball just before it comes in contact with a car, given the height at which it began free fall and the height it hit the car.

To do this, we can use the equation

#(v_y)^2 = (v_(0y))^2 + 2a_y(y - y_0)#

where

  • the final #y#-velocity #v_y# is what we're trying to find

  • the initial #y#-velocity #v_(0y)# is #0#, because it simply dropped on its own to start its free fall

  • the #y#-acceleration #a_y# is #-g#, which is #-9.81# #"m/s"^2#

  • the final position #y# is #2.3# #"m"#

  • the initial position #y_0# is #23.3# #"m"#

Plugging in known values, we have

#(v_y)^2 = (0)^2 - 2(9.81color(white)(l)"m/s"^2)(2.3color(white)(l)"m" - 23.3color(white)(l)"m")#

#(v_y)^2 = (-19.62color(white)(l)"m/s"^2)(-21.0color(white)(l)"m")#

#v_y = sqrt(412.02color(white)(l)"m"^2"/s"^2) = color(red)(20.3# #color(red)("m/s"#

rounded to #1# decimal place.

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