Question

Maury is standing on top of a building. He throws a ball up from the top of the building. The height of the ball is modeled by h(x) = -16t^2 + 12t + 40. How many seconds after release will the ball hit the ground?

A) -2
B) -5/4
C) 5/4
D) 2

Answer 1

The ball will hit the ground 2 seconds after being released by Maury, as determined by solving the quadratic equation representing the ball's height over time and choosing the non-negative root.

Step-by-step explanation:

When Maury is standing on top of a building and throws a ball up, the height of the ball is modeled by the quadratic function h(t) = -16t^2 + 12t + 40. To find out how many seconds after release the ball will hit the ground, we need to solve this equation for when h(t) is equal to zero, which represents the height of the ground. To do this, you must set the equation to zero and factor or use the quadratic formula to find the time t.

In this case, factoring or using the quadratic formula gives us two possible solutions for t. The equation simplifies to the quadratic equation t^2 - (3/4)t - 2.5 = 0. The roots of this equation are t = -5/4 seconds and t = 2 seconds. We can disregard the negative time because it does not fit within our real-world scenario.

Therefore, the ball will hit the ground 2 seconds after Maury releases it, which corresponds to option D).