KINEMATICS: A Leap of Faith

A man leaps off of a 12m building into a large, 4 meter tall vat of quicksand. The man leaves the roof with a horizontal velocity of 18.0 miles per hour. ( 1.000 mi/hr = 0.4470 m/s ).

Q1: A robot on the ground believes that the man is going to miss his mark, and it must move the vat in order to save the man’s life. How much time does the robot have to act before the gentleman misses the top of the vat?

Q2: Where should the center of the vat be located to order to ensure that the man lands safely?

Q3: Initially, the vat is 5m from the wall of the building. Assuming the robot pushes the vat from rest with constant acceleration, what magnitude of acceleration is needed to get the vat positioned correctly in time?

The point of reference is the top of the vat ( 4 m off the ground ). Therefore, the vertical distance ( y ) that is relevant to us is ( y ) = ( 12 m – 4 m = 8 m ).

It is always useful ( if possible ) to determine the total time ( t ) of flight. The man runs off the building and has components of motion in the ( x ) and ( y ) directions that are independent of one another. Nonetheless, the force of gravity determines how long his trip will last. Therefore, an equation relating the vertical distance ( y ) and the time of travel ( t ) is necessary.

( y ) = ( v_oy )( t ) + ( 1 / 2 )( g )( t² ), where ( g ) = gravitational constant of acceleration ( 9.8 m/s² ). The decision to designate the downward ( y ) direction as being positive or negative is arbitrary. Since gravity acts in the downward ( y ) direction, the acceleration due to gravity will be additive to motion in the y-direction. As a consequence, if the downward y-direction is designated as being positive, the ( 1 / 2 )( g )( t² ) term of the kinematic equation must be positive as well.

A1: ( 8 m ) = ( v_oy )( t ) + ( 1 / 2 )( g )( t² ). The initial downward velocity ( v_oy ) component of motion is ( 0 m / s ), so the equation reduces to ( 8 m ) = ( 0 ) + ( 1 / 2 )( g )( t² ). The total time of flight is ( t ) = 1.28 s.

A2: The horizontal distance travelled is ( d ) = ( v_ox )( t ). Using the values given and derived, ( d ) = ( 18.0 mi / h )( 0.4470 m / s )( 1.28 s ) =  10.3 m.

A3: The robot will push the vat outward by a distance of [ ( 10.3 m ) – ( 5 m ) ] = 5.3 m = ( x ). So ( x ) = ( v_ox )( t ) + ( 1 / 2 )( a )( t^2 ). We begin from rest, so ( v_ox ) = 0. Therefore, ( 5.3 m ) = ( 1 / 2 )( a )( t² ). Multiplying both sides of the equation by ( 2 ) gives us ( 10.6 m ) = ( a )( t² ). Squaring the total time of flight ( t ) gives us ( 1.28 s² ) = 1.64 s². Solving for ( a ), we have ( a ) = [ ( 10.6 m ) / ( 1.64 s² ) ] = 6.46 m / s².