A rock is thrown upward with a speed of 48 feet per second from the edge of a cliff 400 feet above the ground. What is the speed of the rock when it hits the ground? Use acceleration due to gravity as –32 feet per second squared and approximate your answer to 3 decimal places.

To find the speed of the rock when it hits the ground, we can use the equations of motion and the concept of conservation of energy.

Let's break down the problem step-by-step:

Step 1: Find the time it takes for the rock to reach the ground.
We can use the equation of motion:
\(y = v_0t + \frac{1}{2}gt^2\), where
\(y\) is the displacement (400 ft, since the rock is 400 ft above the ground),
\(v_0\) is the initial velocity (48 ft/s),
\(g\) is the acceleration due to gravity (-32 ft/s²),
and \(t\) is the time taken.

Plugging in the values, we can rewrite the equation as:
\(400 = 48t - 16t^2\).

Rearranging the equation, we get:
\(16t^2 - 48t + 400 = 0\).

Simplifying the quadratic equation, we can divide everything by 16:
\(t^2 - 3t + 25 = 0\).

Unfortunately, this equation does not factorize nicely, so we'll use the quadratic formula to solve for \(t\):
\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Plugging in the values (a = 1, b = -3, c = 25), we get:
\(t \approx 3.703\) (rounded to three decimal places).

Step 2: Find the final velocity of the rock.
We can use the equation of motion:
\(v = v_0 + gt\),
where \(v\) is the final velocity of the rock.

Plugging in the values, we have:
\(v = 48 + (-32)(3.703)\).

Calculating, we get:
\(v \approx 48 + (-32)(3.703) \approx -18.889\) (rounded to three decimal places).

The negative sign indicates that the velocity is directed downward.

Step 3: Find the speed of the rock when it hits the ground.
The speed is the magnitude of the velocity, so we need to take its absolute value:
Speed = |v| = |-18.889| = 18.889 ft/s (rounded to three decimal places).

Therefore, the speed of the rock when it hits the ground is approximately 18.889 ft/s.

a=dv/dt

v=INT a dt=-32t + k
but at t=0 v=48 so k=48
v=-32t+48
h= INT v dt= int( -32t +48)dt
h=-16t^2 + 48t + k2
but at h(400), at t=0, this means k2 is 400

h=-16t^2+48t + 400
when h=0, find t
16t^2-48t-400=0
t^2-3t-25=0

t=(3+-sqrt(9+100))/2
t=1.5-+sqrt(109/4)) use positive number...
now find v at that time...
so what is v when