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)

Question

Please help me solve this

Answer 1

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

Answer 2

On the 9th second its 450m

I hope this helps :-)

Answer 3

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

Answer 4

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.

Answer 5

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.

Answer 6

if d = k t^2
You found k = 5
to find distance in 9 th second find d (9) and d(10) and subtract
d(10) = 5 (100)
d(9) = 5(81)
so answer = 5 (100 -81) = 5*19 = 95

Answer 7

Hope that helped :)

the first second is between t = 0 and t = 1
the second second is between t = 1 and t = 2
the third second is between t = 2 and t = 3

etc
so the ninth is between t = 9 and t = 10

Damon Cummings