Question

The height h(t) at the time t of a rock tossed into the air at time 0 from an initial height h₀ at an initial velocity V₀ is given by h(t) = h₀+ v₀t – 16t². where time is measured in seconds, height in feet, and velocity in feet per second. The positive direction is up, so if the rock is moving down then its velocity is negative. The magic number 16 in this equation is due to the mass and radius of earth and would be different for example on Mars or on the Moon. Suppose you throw a rock upward from a height of 64 feet with an initial velocity of 48 feet per second. The rock will hit the ground after ______seconds.

Answer 1

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.