Final answer:
To determine when a rock thrown upward will hit the ground, we solve the quadratic equation h(t) = h₀ + v₀t − 16t² by setting h(t) to 0 and finding the positive value of t. After substituting the given values and factoring, we find that the rock will hit the ground after 4 seconds.
Step-by-step explanation:
The student asks how long it will take for a rock thrown upward from a height of 64 feet with an initial velocity of 48 feet per second to hit the ground. The equation given is h(t) = h₀ + v₀t − 16t², where h(t) represents the height of the rock at time t, h₀ is the initial height, and v₀ is the initial velocity. To find out when the rock will hit the ground, we need to solve for t when h(t) = 0 (the height is zero).
Substituting the known values into the equation gives us:
0 = 64 + 48t − 16t².
Rearranging terms to set the equation to standard quadratic form:
16t² − 48t − 64 = 0.
Then we would divide the entire equation by 16 to simplify it:
t² − 3t − 4 = 0.
Now, we can factor the quadratic equation or use the quadratic formula to find the values of t.
Factoring gives us:
(t − 4)(t + 1) = 0.
We find two potential solutions for t: t = 4 seconds and t = − 1 second. Since time cannot be negative, we disregard t = −1. Thus, the rock will hit the ground after 4 seconds.