Q: A model-rocket club is standing at the edge of a 40.0 m building. They arrange a launcher so that it is off the cliff’s edge, and they launch a rocket straight upward at an initial velocity ( v ) of 380.0 m/s. Neglecting wind resistance, how long will it take for the rocket to fall to the earth’s surface?
A: As we will soon see, this question requires the usage of several kinematic equations. Prior to identifying which equations are needed, a brief discussion about assigning positive and negative signs to our variables is warranted. Let’s begin by looking at a diagram of the proposed path to be traveled:
The most important step is to ensure that the gravitational ( g ) constant of acceleration is given an appropriate sign, whether it be positive or negative. It is useful to think of the force of gravity ( Fg ) as being like a river that is heading downward. If the upward direction is assigned a positive value, anything that travels in this direction will be going against the river; therefore, the ( g ) constant must be assigned a negative value. Once initial positive and negative values are assigned to the variables in question, they must NOT be changed throughout the course of working the problem. If this is done, the mathematics will end up being incorrect. Let’s now look at two other possible scenarios to further solidify the point being made.
Let’s assume that, for the sake of convenience, the downward direction is assigned a positive value in a separate exercise. In this scenario, anything that moves downward will be moving with the river. As a consequence, the ( g ) constant must also be given a positive sign designation.
Finally, what sign will be assigned to ( g ) if the down direction is assigned a negative value??? Actually, this is the scenario we’ve been presented with in the question at hand. Please recall that the upward direction has been assigned a positive value in the scenario that is given. When the model rocket reaches its apex and subsequently falls back to its original height above ground, it will be moving at a negative 380.0 m/s due to the symmetric nature of projectile motion. At this point, the rocket will be moving with the river, but if we’d assigned a positive value to ( g ), the physics of the system wouldn’t make much sense.
In order to begin solving the problem, we must first determine how much time ( t ) transpires as the rocket launches and travels to its apex of flight. Notice, however, that we do not know how high the rocket will fly, and we haven’t been asked to make this determination, either. For this reason, we must first identify an equation that relates the velocity ( v ) of the rocket in the +y-direction to the gravitational ( g ) constant of acceleration:
vf1 = vi1 + a1t1
a1 = g
vf1 = vi1 + gt1
Hold up!!!
Since the upward direction has a positive value assigned to it, the ( g ) constant must be negative:
vf1 = vi1 – gt1
We must also realize that the final velocity will be equal to zero when the rocket reaches its apex of travel. This information enables us to simplify the equation and solve for ( t ):
0 = vi1 – gt1
– vi1 = – gt1
t1 = ( – vi1 / – g ) = ( vi1 / g )
Well, this is almost the expression that we want; it will tell us how long it took for the rocket to reach its apex, but we need to know how long it takes for the rocket to return to its original level above ground. If we neglect wind resistance, the rocket moves upward and downward over equal time intervals; therefore, the round trip, thus far, is quantified with the following expression:
2t1 = ( 2vi1 / g )
We must now determine how much time the rocket spent in free-fall after passing its launch height moving downward. This amount of time will be less than it would if the rocket had simply been dropped from the height in question, because the initial velocity ( – vi2 ) on the way downward is not zero:
– vf2 = – vi2 – gt2
We know the value of ( – vi2 ) as well as the value of ( g ), so in order to solve for ( t ), we must use another equation to solve for ( – vf2 ). We know that the height in question will influence the amount of time the rocket spends falling to earth, and we’ve been given the launch height. Let’s now use the kinematic equation that relates velocity, distance, and acceleration ( or deceleration ) to one another:
( – vf2 )2 = ( – vi2 )2 + 2( – g )( – y )
( – vf2 )2 = ( – vi2 )2 + 2gy
v2f2 = v2i2 + 2gy
Note: Remember, we assigned a positive value to the upward ( y ) direction; thus, the variables must now reflect the fact that we’re heading in a downward direction. If we wanted, we could have just divided the trip into two parts: ( 1 ) where the rocket moves upward, stops, and reverses direction until reaching its launch height, and ( 2 ) the trip downward from the launch height to the earth’s surface. For the sake of mathematical convenience, both velocity terms ( v ), along with the ( g ) and ( y ) terms, could be given positive values. By doing so, we end up with the exact same equation is we would have derived above:
v2f2 = v2i2 + 2gy
Solving for the time that transpired over the duration of the trip is now a matter of substitution of known values into our equations:
t total1 = 2t1 = ( 2vi1 / g )
2t1 = [ ( 2 )( 380.00 m/s ) / 9.80 m/s2 ]
t total1 = 77.6 s
– vf2 = – vi2 – gt2
– vf2 = – ( vi2 + gt2 )
vf2 = vi2 + gt2
vi2 = 380.0 m/s
g = 9.80 m/s2
vf2 = 380.0 m/s + ( 9.80 m/s2 )t2
Solving for ( v2f2 ) will enable us to solve for ( t2 ):
v2f2 = v2i2 + 2gy
v2f2 = ( 380.0 m/s )2 + 2( 9.80 m/s2 )( 40.0 m )
v2f2 = 144,400 m2/s2 + 784 m2/s2
v2f2 = 145,184 m2/s2
vf2 = 381.0 m/s
381.0 m/s = 380.0 m/s + ( 9.80 m/s2 )t2
1.0 m/s = ( 9.80 m/s2 )t2
t total 2 = 0.102 s
t total = t total1 + t total 2
t total = 77.6 s + 0.102 s
t total = 77.7 s
RENAISSANCE PHYSICS 