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*) = -4.9

*t*

^{2}+ 19.6

*t*+ 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 = -4.9t^{2}+ 19.6t+ 58.8

0 =t^{2}− 4t− 12

0 = (t− 6)(t+ 2)t= 6 ort= -2

*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.

**6 seconds**after launch.

**Formula of Projectile Motion Problems**

*s*(

*t*) = -

*gt*

^{2}+

*v*

_{0}

*t*+

*h*

_{0}

is the height at any particular time (*s**t*) [Note:*s*(*t*) is also sometimes shown in the formula as*h*]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.]**g**is the initial velocity*v*_{0}is the initial height*h*_{0}

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*) = -*g*^{2} + *v*_{0}*t* + *h*_{0} **Step 2:** Use the gravity value in meters, insert the initial velocity and heights given: *s*(*t*) = -4.9*t ^{ }*

^{2}+ 39.2

*t*+ 0

**Step 3:**We need the times when the height

*s*(

*t*) = 34.3 m

**Step 4:**Solve -4.9

*t*

^{ }^{2}+ 39.2

*t*= 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?

*s*(

*t*) = -16

*t*

^{ }^{2}+ 160

thrown object:

*s*(

*t*) = -16

*t*

^{ }^{2}− 48

*t*+ 160

*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