solve the equations by factoring

Question

An object is thrown downward, with an initial speed of 16 ft/s, from the top of a building 480 ft high. How many seconds later will the object hit the ground? Use the equation

d = vt + 16t2,
where d is the distance in feet, v is the initial speed, and t is the time in seconds.


 

Answer 1

So, we are given the formula  d = vt + 16t² for the distance the object travels after it is thrown. To get started, we have to realize that the object will hit the ground after it has travelled 480 feet (the height of the building). This tells us that d = 480, and we are given that v = 16. We are trying to determine the value of t that will make the equation true.

 

Putting the numbers into the formula, we have  480 = 16t + 16t², and we are asked to solve it by factoring. To do this, one side of the equation must be 0, so we subtract 480 from both sides to make the left size 0.

 

This gives us  0 = 16t + 16t² - 480, which we can rearrange to get

 

16t² + 16t - 480 = 0

Don't forget that the first step of every factoring problem is to factor out the GCF, if there is one. In this case, there is a GCF of 16, which we can factor out to get

 

16(t² + t - 30) = 0. We can divide both sides by 16, since it won't affect the solutions, giving us   t² + t - 30 = 0.

To factor this, we need two numbers whose product is -30 and whose sum is 1. That would be  6 and  -5. Using these numbers, we factor the problem into  (t+6)(t-5)=30. Always stop to multiply it out, to make sure you have factored correctly. If you haven't, you can't get the right answers.

 

Setting each factor equal to 0, we have   t+6=0  or  t-5=0, leading to solutions  t = -6   or   t = 5.

 

What? It hits the ground at two different times?

No, the -6 time is referring to before the object was thrown, so it can be disregarded.

The object hits the ground 5 seconds after it was thrown.