Q: A mass m is moving horizontally across the surface of the earth with a velocity of ( v ) = 30 m/s . The mass approaches a ledge with an edge situated 100 m above the ocean. The mass eventually moves off the ledge and falls until it reaches the ocean’s surface. Calculate the furthest horizontal distance reached by the mass.
A: Several considerations must be made in order to address the given circumstance. First and foremost are considerations regarding unbalanced forces ( if any ) that act upon the mass after it moves off of the ledge. Unbalanced forces cause objects that are “ at rest “ ( or initially moving with a constant velocity ) to move into a state in which its velocity is constantly changing ( accelerating ) with the passage of time. Such an object will continue to accelerate until an equal and opposite force opposes the force that put it into motion. When ( or if ) an equilibrium force pair is established, an object will move at a constant velocity. If the object was initially moving and brought to rest, it will remain at rest until it is acted upon by another force. Velocity ( v ) is quantified in units of meters per second ( m / s ), and acceleration is quantified by how much an object’s velocity changes in ( m / s ) every second ( s ). Thus, the units of linear acceleration ( a ) are in [ ( m / s ) / s ] or ( m / s^2 ). Other types of acceleration ( Ex. centripetal ) exist, but the equations that quantify “ center seeking “ or periodic motion are not applicable here.
Before moving off the ledge, the mass moves with a constant velocity of 30 m/s in the horizontal direction. In this problem, wind resistance is negligible. For this reason, the mass will move with a constant velocity in the x-direction. Since no forces act upon the mass in the x-direction, the mass will continue travelling in this direction until it lands in the ocean; however, the earth will begin to exert a gravitational force of attraction upon the mass in the vertical y-direction as it leaves the ledge. The formula for the gravitational force of attraction between two masses is F = Gm1m2/r^2. When masses much smaller than the earth are near the earth’s surface, the values of m1 ( mass of the earth ), the constant ( G ), and the earth’s radius ( r ) are constant, and the force equation above reduces to F = ( Gm1/r^2 )( m2 ) = mg. In the kinematic equation of motion that follows, the acceleration due to gravity ( g ) = 9.8m/s^2 will be used to quantify the influence of the earth’s gravitational pull upon the mass in the y-direction.
The distance the mass will travel in the x-direction is equal to d = vt, where distance ( d ) is in meters ( m ), velocity is in meters per second ( m / s ), and t = the total time of flight in seconds ( s ). Since only the velocity of the mass is given ( 30 m/s ), we must somehow or another relate the total time of flight ( t ) to the height of the ledge ( y ). Fortunately, the kinematic equation of y = vot +/- 1/2at^2 relates distance travelled in the vertical direction to the initial velocity ( if any ) and acceleration of an object in free fall.
The top of the ledge is the point of reference, and the direction downward can be designated as being either positive or negative. The choice of sign designation regarding distance will determine whether the positive ( + ) or negative ( – ) sign designation is used in the latter half of the equation. If downward is designated as being negative ( – y ), the bottom of the ledge will be a distance of -100m from the top. Thus, the +/-1/2at^2 term must become increasingly negative over time to make any mathematical sense. As a consequence, the appropriate equation is -y = -vot – 1/2at^2.
Fortunately, the initial velocity in the -y direction ( vo ) = 0. As a consequence, the equation reduces to -y = -1/2at^2. Recall that the acceleration in the y-direction is due to g = 9.8m/s^2. Therefore, “ g “ will replace “ a “ in the equation above. Using -y = -1/2gt^2, we can now derive an equation that will allow us to solve for the total time of flight ( t ).
First, ( -y )( -2 ) = gt^2. Notice how the negative signs cancel? This re-emphasizes the importance of having chosen a ( + ) or ( – ) value of +/-1/2at^2 that mathematically corresponds with the vertical direction travelled as being negative. We now have 2y = gt^2. Solving for total time of flight ( t ), we have the square root of 2y/g = t, rewritten as ( sqrt 2y/g ) = t. Since 2y = 200m, and g = 9.8m/s^2, the ( sqrt 200m/ 9.8m/s^2 ) = 4.52s. Finally, when we substitute appropriate values into d = vt, the maximum distance the mass will travel is d = ( 30m/s )( 4.52s ) = 135.6 m ( close to 140 m ).
In any circumstance where an unbalanced force is acting upon an object in one direction as it moves with a constant velocity in another direction, the curvilinear motion of the object justifies use of a uniform acceleration equation. If the motion vectors of the object were equal in magnitude, the object would move diagonally in its descent into the ocean. If the gravitational force of attraction were altogether absent, the use of a uniform acceleration equation would have been unnecessary; absent the gravitational force of attraction exerted by the earth, the mass would travel indefinitely in the x-direction until acted upon by another force.
RENAISSANCE PHYSICS 