Question

A ball is dropped on to the floor from a height of 10 m. It reboundes to a height of 2.5 m. If the ball is in contact with the floor for 0.01 seconds, what is the average acceleration during contact?

a) \[1400 m/s^2\]

b) \[2100 m/s^2\]

c) \[700 m/s^2\]

d) \[2800 m/s^2\]

Answer 1

@amistre64 @hba @asnaseer @AccessDenied @jhonyy9 @jazy @ganeshie8

Answer 2

The acceleration of a body is defined as the change of its velocity during an interval of time divided by the interval of time. Expressed algebraically, a = (Vo - Vf)/t where a = the acceleration, Vo = the initial velocity, Vf = the final velocity and t = the interval of time.

Answer 3

From the basic a = (Vo - Vf) we derive the first as
......................................t
..............................Vf = Vo + at (the acceleration is assumed constant)

Answer 4

The second defines the relationship between the distance traveled by a body exposed to a constant acceleration. The distance traveled over an interval of time derives from the product of the average velocity during the time interval and the time interval. The average velocity is (Vo + Vf)/2 and the time interval "t" which yields d = (Vo + Vf)t/2. Substituting the Vf derived above, d = [Vo + (Vo + at)]t/2 resulting in

...............................d = Vo(t) + a(t^2) (quite often, s is used in place of d)
......................................................2

Answer 5

The third combines the first two by eliminating the time interval "t".
Therefore, a(d) = (Vo - Vf)(Vo + Vf)t = (1/2)(Vf - Vo)(Vf + Vo) resulting in
.....................................t.............2

..............................Vf^2 = Vo^2 + 2ad

Answer 6

thnx.... @tictac4456

Answer 7

no problem do u go to connexus

Answer 8

is it right??


While falling down, initial velocity = 0, final velocity = v, acceleration due to gravity = 9.8 m/s 2 , height = 10 m
The velocity of the ball with which it hits the ground can be found using,
v 2 = u 2 + 2as
=> v 2 = 0 + 2×9.8×10
=> v = 14 m/s (downward)
While it moves up, initial velocity = u, final velocity = 0, acceleration due to gravity = -9.8 m/s2 [this time the velocity and acceleration are oppositely directed so we have the negative sign] , height = 2.5 m
The velocity of the ball with which it leaves the ground can be found using,
v 2 = u 2 + 2as
=> 0 = u 2 - 2×9.8×2.5
=> u = 7 m/s (upward)
So, Change in velocity = 7 m/s (upward) – 14 m/s (downward)
=> Δv = 7 m/s + 14 m/s [both in upward direction]
=> Δv = 21 m/s
So, acceleration is, a = Δv/Δt = 21/0.01 = 2100 m/s 2

Answer 9

yup

Answer 10

@tictac4456

Answer 11

thats correct

Answer 12

thnx..

Answer 13

@satellite73 help me:)

Answer 14

@amistre64

Answer 15

hey ur work is correct but the answer is wrong

Answer 16

the answer is A

Answer 17

howz that??

Answer 18

oh im srry never mind i didnt see the seconds ur answer is correct

Answer 19

im 100% sure its right

Answer 20

ohkkk!!

Answer 21

I agree - your work and answer is correct

Answer 22

thnx...

Answer 23

yw :)

Answer 24

do ya take entrepuener class

Answer 25

what do you mean??

Answer 26

other than physics do u take entrepuener class

Answer 27

its like a business class

Answer 28

no!! never!!! why??

Answer 29

haha
i kinda needed help

Answer 30

but never mind

Answer 31

plz you can!!!

Answer 32

ok lol but if u dont knw it u dont have to help k

Answer 33

k

Answer 34

what is gross profit?

A)the money left over after the cost of making a product or providing a service
B) the cost of labor to produce a product
C) the difference between a product's price and the revenue of the company
D) a negative net profit, in which company experiences a loss

Answer 35

(finance) the net sales minus the cost of goods and services sold.

Answer 36

thnkz
lol

Answer 37

i gave u a medal right

Answer 38

right!!!

Answer 39

cool so did u graduate? or u still in high skool

Answer 40

high school!!!

Answer 41

what grade

Answer 42

9 class

Answer 43

ok @tictac4456

Answer 44

me2 lol

Answer 45

what skool?