Quadratic Application: Projectile Motion


Quadratic Applications: Projectile Motion 

Generally speaking, projectile motion problems involve objects that are thrown, shot, or dropped. Usually the object will be launched directly upward or dropped directly down.


Consider the following example:

An object is launched directly upward at 19.6 m/s from a 58.8-meter tall platform.  The equation for the object's height (s) at time (t) seconds after launch is s(t) = -4.9t2 + 19.6t + 58.8, where s is in meters.  When does the object strike the ground?

Understand

This question asks for the time when the object strikes the ground.  Another way of thinking of this is, when is the object's height zero (on the ground)?  So, set s equal to zero and solve the equation:

Solve
0 = -4.9t2 + 19.6t + 58.8
0 = t2 − 4t − 12
0 = (t − 6)(t + 2)
t = 6 or t = -2

Recall that t is a value of time, answering the question "when."  It doesn't make sense in this context for t to be a negative value, so t = -2 is an extraneous solution.

The object strikes the ground 6 seconds after launch.

Formula of Projectile Motion Problems

The key features to know in projectile motion problems are the initial height, the initial speed, a
value of the force of gravity, and time.  They are related in the formula given below:

s(t) = -gt2 + v0t + h0

where:

  • s  is the height at any particular time (t)  [Note:  s(t) is also sometimes shown in the formula as h]
  • g is gravity value — in feet this value is 16 and in meters this value is 4.9  [Note:  In physics, the gravitational constant is actually 32 for feet and 9.8 for meters, but the formula uses one-half this value.]
  • v0 is the initial velocity 
  • h0 is the initial height




Practice

1)  An object is launched from ground level directly upward at 39.2 m/s.  For how long is the object at or above a height of 34.3 meters?

Step 1:  Recall the formula:  s(t) = -g2 + v0t + h0  

Step 2:  Use the gravity value in meters, insert the initial velocity and heights given:  s(t) = -4.9t 2 + 39.2t + 0 

Step 3:  We need the times when the height s(t) = 34.3 m

Understand:  Applying the formula in this way will yield the two times when the object is exactly 34.3 meters high.  The first time it reaches that height it is on its way up; the second time it reaches that height it is on its way down.  The difference between the two times represents the length of time the object is at or above the height of 34.3 meters.

Step 4:  Solve  -4.9t 2 + 39.2t = 34.3 and subtract the two answers.






2)  Two objects are dropped from a bridge 160 feet above ground level.  One object is released with an initial velocity of zero, falling only by the force of gravity.  The other is thrown straight down with a velocity of 48 ft/s.  How many seconds earlier does the thrown object reach the ground as compared with the dropped object?

dropped object:  s(t) = -16t 2 + 160
thrown object:  s(t) = -16t 2 − 48t + 160

Understand:  Find the time when the height is zero [s(t) = 0] for each object and find the difference in times.









3)  An object is thrust directly down from a height of 190 feet at a velocity downward of 64 ft/s.  How many seconds will it take before the object hits the ground?



















Answers

1)  6 seconds 

2)  1.16 seconds

3)  5.98 seconds