Question

The height, h, of a basketball about the ground (in feet) is given by the formula

h = −32t2 + 160t, where t is the number of seconds since the ball was thrown. How many seconds after it was thrown does it take for the ball to land?

Answer 1

The ball will land when h = 0, so we can solve for t by setting the formula equal to 0 and solving for t:

-32t^2 + 160t = 0

Factor out a t:

t(-32t + 160) = 0

Solve for t:

t = 0 or -32t + 160 = 0

The solution t = 0 corresponds to when the ball is first thrown, so we can ignore it. Solving for -32t + 160 = 0 gives:

-32t = -160

t = 5

Therefore, the ball will land 5 seconds after it was thrown.

Step-by-step explanation:

Answer 2

Answer: The ball lands 5 seconds after it was thrown.

Step-by-step explanation:

The basketball hits the ground when h = 0. So, we set h = 0 in the equation and solve for t:

0 = -32t² + 160t

This is a quadratic equation in the form of at² + bt + c = 0. We can factor out a common factor of -32t:

0 = -32(t - 5)

Setting each factor equal to zero gives the solutions to the equation:

-32t = 0 => t = 0

t - 5 = 0 => t = 5

So, the times when the ball is on the ground are t = 0 (when it was first thrown) and t = 5 seconds (when it lands). Therefore, the ball lands 5 seconds after it was thrown.