Create an image portraying the abstract concept of motion influenced by gravity, represented through a sphere with lines representing motion, tracing an arc into the sky. Illustrate the curvature revealing the effects of speed reduction over time, yet show no numbers or equations to ensure no text. The scene is situated on Earth with grass and blue sky for reference. All elements must convey the matter of physics, specifically the transformation of kinetic energy under the effect of gravity and the drop in velocity.

A body rises vertically from the earth according to the law s = 64t – 16t2. If it has lost k times its velocity in its 48 ft rise, then k =

v = ds/dt = 64 - 32t 

at t = 0 , v = 64 
when s = 48 
48 = 64t - 16t^2 
t^2 - 4t + 3 = 0 
(t-1))(t -3) = 0 
t = 1 or t = 3 
at t = 1, it is still rising 
v = 64 - 32= 32

So 64k=32
K= 1/2

v = ds/dt = 64 - 32t

at t = 0 , v = 64
when s = 48
48 = 64t - 16t^2
t^2 - 4t + 3 = 0
(t-1))(t -3) = 0
t = 1 or t = 3
at t = 1, it is still rising
v = 64 - 16 = 48

so 64k = 48
k = 48/64 = 3/4

Well, let's find out! We can start by finding the velocity of the body at the end of the 48 ft rise.

Given that the equation for velocity is v = ds/dt, we can differentiate the equation for s to find the velocity function:

v = d(64t - 16t²)/dt
v = 64 - 32t

To find the velocity at the end of the 48 ft rise, we substitute s = 48 into the equation for s:

48 = 64t - 16t²
16t² - 64t + 48 = 0

Simplifying this quadratic equation, we get:
t² - 4t + 3 = 0
(t - 1)(t - 3) = 0

So, t = 1 or t = 3. Since we're only interested in the end of the 48 ft rise, we'll use t = 3.

Now, let's find the velocity at t = 3:
v = 64 - 32t
v = 64 - 32(3)
v = 64 - 96
v = -32

So, the velocity of the body at the end of the 48 ft rise is -32 ft/s.

Now, let's find out how many times this velocity it has lost.

To do that, we divide the initial velocity by the velocity it has lost:
k = 64 / 32
k = 2

Therefore, k = 2.

To find the value of k, we need to determine the velocity of the body during its 48 ft rise.

The velocity of an object is the derivative of its displacement function. In this case, the displacement function is given as s = 64t - 16t^2. Let's find the derivative:

ds/dt = 64 - 32t

Now, we can find the velocity of the body when its displacement is 48 ft. We set s = 48 and solve for t:

64t - 16t^2 = 48

Rearranging the equation, we get:

16t^2 - 64t + 48 = 0

Dividing the equation by 16, we have:

t^2 - 4t + 3 = 0

The equation can be factored:

(t - 1)(t - 3) = 0

So, t = 1 or t = 3.

Now, let's substitute t = 1 into the velocity equation ds/dt = 64 - 32t:

ds/dt = 64 - 32(1) = 64 - 32 = 32 ft/s

Therefore, the velocity of the body at a displacement of 48 ft is 32 ft/s.

Next, we need to find k, which represents the number of times the body has lost its velocity during the 48 ft rise.

We know that velocity is lost when it becomes zero. So, let's set the velocity equation equal to zero and find the time when the velocity is zero:

64 - 32t = 0

Solving for t, we get:

32t = 64

t = 64/32

t = 2 seconds

So, the velocity of the body becomes zero at t = 2 seconds.

To find k, we divide the time it takes for the velocity to become zero (2 seconds) by the time it takes for the body to rise 48 ft (t = 1 second):

k = 2/1 = 2

Therefore, k = 2.

To find the value of k, we need to first determine the velocity of the body at any given time during its rise.

The equation given is s = 64t – 16t^2, where s represents the height (in feet) and t represents the time (in seconds).

The velocity of an object is the derivative of its position with respect to time. In this case, the derivative of the position s with respect to time t gives us the velocity v.

Taking the derivative of the equation s = 64t – 16t^2, we get:

v = ds/dt = d/dt(64t – 16t^2)

To find the derivative, we can apply the power rule for differentiation. The power rule states that if we have a term of the form ax^n, where a and n are constants, the derivative is given by:

d/dt(ax^n) = nax^(n-1)

Applying the power rule, we differentiate each term in the equation:

v = 64 * d/dt(t) - 16 * d/dt(t^2)
v = 64 * 1 - 16 * 2t
v = 64 - 32t

Now we have the velocity as a function of time: v = 64 - 32t.

The question states that the body has lost k times its velocity during its 48 ft rise. This means that at some point, the velocity will decrease to k/1 times its initial velocity.

Let's assume that the initial velocity of the body is v0. So at the time t when the body has lost k times its velocity, the new velocity will be v0/k.

We can now solve for k. Set v = v0/k:

v0/k = 64 - 32t

To find the value of k, we need an additional condition or value. Without further information, it is not possible to determine the exact value of k.