Question

If an object is propelled upward from a height of 144 feet at an initial velocity of 128 feet per second, then its height after t seconds is given by the equation h(t)=-16t²+128t+144. After how many seconds does the object hit the ground?

a. 9 Seconds
b. 8 Seconds

Answer 1

To determine after how many seconds the object hits the ground, solve the quadratic equation h(t) = -16t² + 128t + 144 for when h(t) is zero. The solution is 9 seconds, which is when the object will hit the ground.

Step-by-step explanation:

The student's question involves determining the time when an object, propelled upward with an initial velocity from a certain height, hits the ground. The equation provided, h(t) = -16t² + 128t + 144, represents the height of the object as a function of time where 't' is the time in seconds and 'h(t)' is the height in feet.

To find out after how many seconds the object hits the ground, we need to solve for 't' when the height 'h(t)' is zero. Setting the equation to zero gives us the quadratic equation 0 = -16t² + 128t + 144. To solve this equation, we can either factor it directly, complete the square, or use the quadratic formula.

In this case, factoring is possible, and we find the equation factors to -16(t - 9)(t + 1). Since time cannot be negative in this context, we disregard the solution 't = -1' and take the positive solution, 't = 9'. Therefore, the object will hit the ground after 9 seconds.