Illustrate a physics-themed scenario: A body falling freely from a significant height in a clear sky, creating a sense of motion. In the background, depict the outlines of a stopwatch showing 9 seconds, and a meter scale which is divided into segments, marks till 500 m visible. The body in depiction should be geometrical in shape (sphere or cube), to exclude any possible human-like features.

the distance that a body falls through when dropped from a certain height varies directly with the square of the time of fall. A body falls through a total of 500m in 10 seconds. find the distance it falls through, in 9th second (in m)

Please help me solve this

its 85 metres

Sol
first find for 9 seconds
500 = k * 10 * 10 => k=5
for 9 seconds
D= 5 * 9 * 9 = 405 so 95 seconds travelled
for 8 seconds D = 5 * 8 * 8 = 320
1 second it travells 95 metre so 9 th second 320 + 95 = 415
so in 9th second it travells exactly 500 - 415 = 85.

Ans - 85 metres

Sol
first find for 9 seconds
500 = k * 10 * 10 => k=5
for 9 seconds
D= 5 * 9 * 9 = 405 so 95 seconds travelled
for 8 seconds D = 5 * 8 * 8 = 320
1 second it travells 95 metre so 9 th second 320 + 95 = 415
so in 9th second it travells exactly 500 - 415 = 85.

Well, let's dive into this problem, shall we? The distance that a body falls varies directly with the square of the time of fall. So, we can form an equation to represent this relationship.

Let's call the distance it falls "d" and the time of fall "t". According to the problem, when the body falls for 10 seconds, it covers a distance of 500 meters. So, we can set up the following equation:

d = kt^2

Now, we need to find the value of "k", the constant of variation. To do this, we can substitute the values given into the equation. When t = 10, d = 500. Plugging these values into the equation, we get:

500 = k * 10^2
500 = 100k

Now, we can solve for k:

k = 500 / 100
k = 5

So, the equation for the distance fallen is d = 5t^2.

Now, we need to find the distance the body falls through in the 9th second. To do this, we substitute t = 9 into our equation:

d = 5 * 9^2
d = 5 * 81
d = 405 meters

Therefore, the distance the body falls through in the 9th second is 405 meters. Hope that tickled your mathematical funny bone!

To solve this problem, we need to use the law of falling bodies, which states that the distance a body falls is directly proportional to the square of the time of fall.

Let's first define the variables:
d - distance fallen
t - time of fall

According to the problem, we know that the distance the body falls through varies directly with the square of the time of fall. This can be expressed as:

d = kt^2

Where k is the constant of proportionality.

We also know that the body falls through a total of 500m in 10 seconds. So we have:

500 = k(10^2)

Simplifying this equation, we get:

500 = 100k

Divide both sides by 100 to solve for k:

k = 500/100
k = 5

Now, we can use the equation d = kt^2 to find the distance the body falls in the 9th second:

d = 5(9^2)
d = 5(81)
d = 405

Therefore, the body falls through a distance of 405m in the 9th second.

sorry...can't help you on this.

On the 9th second its 450m

I hope this helps :-)

I can help . . . D varies t^2 . Which means d=kt^2. 500 =10^2 x k . 500 = 100k. K =constant =5 . The law goes lyk dis. D =5t . But we must nt neglect aceleration due to gravity=10ms^2. D =5 x 9 x 10=450m. . . Ansa =450* coz when 10s =500m . 9s = 450m