if a basketball is projected upward from ground level with a velocity of 64 feet per second its high(y) can be predicted from the number of seconds the object is in the air (x) using the function

Question

How high will the ball be after 1 second?

How many seconds will it take for the ball to reach its maximum height of 64 feet?

How many seconds does it take overall fr the ball to return to the ground?

Answer 1

Hi Angie

The question you posted omitted the function. It should be something like

y = 64 x - 16 x^2. where y is the height, and x is the time of flight.

To answer the questions: 1) The height of the ball after 1 sec is the value of y when x is equal to 1, so it's

64 *1 - 16 * 1 = 48 ft

2) At it's maximum height , y = 64; so your equation looks like 64= 64x -16x^2

This is a quadratic, which can be reorganized thus: 16x^2 -64x +64 = 0

Dividing by 16 yields x^2-4x+4 = 0 Factoring : (x-2)(x-2) = 0 So the time at the peak is x = 2 sec

3) To hind the time it takes to return to earth, note that when it lands, y = 0

So, the equation this time is 0 = 64 x - 16 x^2 Divide both sides by 16 and rearrange x^2 = 4 x

This equation has two solutions x=4, and x=0. x=0 corresponds to the position before the ball is thrown

and x=4 is when it lands. Note that going up and coming down takes the same amount of time.