Word problem

Question

Suppose you throw a ball straight up from the groin with a velicity of 80 ft/s. As the ball moves upward, gravity slows it. Eventually the ball begins to ball back to the ground. The height h of the ball after t seconds in the air is given by the quadratic finction h(t) = -16t² + 80t

 

a) how high does the ball go?

b) for how many seconds is the ball in the air before it hits the ground? 

Answer 1

Let's answer "b" first.

 

The ball hits the ground when h(t) = 0

 

h(t) = -16t² + 80t

0 = -t(16t - 80)

 

If the product of two numbers is zero, one of the other number must be zero.

 

-t = 0, This is when the ball is thrown.

16t - 80 = 0

16t = 80

t = 5

 

This is when the ball hits the ground.

 

Now how high does the ball go? It comes down in the same amount of time that it goes up.

 

When t = 2.5, it is as high as it can go.

 

h(2.5) = -16t² + 80t

h(2.5) = -16(2.5)² + 80(2.5)

h(2.5) = -16(6.25) + 200

h(2.5) = -100 + 200

h(2.5) = 100

 

The ball reaches 100 feet.