Question

an object is dropped from an initial height of 1024 feet. in how many seconds will it reach the ground? use a graph to help you solve.

Explanation:

Step1: Recall the free - fall formula

The height formula for an object in free - fall is $h(t)=h_0+v_0t-\frac{1}{2}gt^2$, where $h_0$ is the initial height, $v_0$ is the initial velocity, $t$ is the time, and $g$ is the acceleration due to gravity. Since the object is dropped, $v_0 = 0$. In English units, $g = 32$ ft/s² and $h_0=1024$ feet. So the equation becomes $h(t)=1024 - 16t^2$.

Step2: Set $h(t)=0$

We want to find the time $t$ when the object reaches the ground, so we set $h(t)=0$. Then $0 = 1024-16t^2$.

Step3: Solve for $t$

First, rewrite the equation as $16t^2=1024$. Then divide both sides by 16: $t^2=\frac{1024}{16}=64$. Taking the square root of both sides, we get $t = 8$ (we ignore the negative root since time cannot be negative in this context).

Answer:

8 seconds