Question in description

Question

 A basketball player shot a basketball from 6  feet high at a speed of 20 feet per second. The height of the ball off the ground is given by the formula h(t)=-16t^2+20t+6, where h(t) represents the ball's height at time ,t, seconds after the ball is thrown. The downward negative pull on the ball is represented by -16t^2 When does the ball hit the ground?

A)1.5 seconds

B)1.75 seconds

C)1.25 seconds

D)0.25seconds

Answer 1

Set h(t) to zero and solve for t.

0 = -16t^2 + 20t + 6

Multiply both sides by negative one.

0 = 16t^2 - 20t - 6

Divide both sides by two.

0 = 8t^2 - 10t - 3

Factor this quadratic equation.

0 = (4t + 1)(2t - 3)

So,

4t + 1 = 0 --or-- 2t - 3 = 0

4t = -1 --or-- 2t = 3

t = -1/4 --or-- t = 3/2

Since I haven't found where the Doctor has parked the TARDIS, we can't go back in time, so we must eliminate t = -1/4. Therefore t = 3/2 = 1.5 seconds.

Answer 2

The ball hits the ground at a height of 0 (zero).

0 = -16t² + 20t + 6

Can you solve for t and answer?