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mayankdevnani
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) \[2800 m/s^2\]
@amistre64 @hba @asnaseer @AccessDenied @jhonyy9 @jazy @ganeshie8
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.
From the basic a = (Vo - Vf) we derive the first as ......................................t ..............................Vf = Vo + at (the acceleration is assumed constant)
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
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
thnx.... @tictac4456
no problem do u go to connexus
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
@satellite73 help me:)
hey ur work is correct but the answer is wrong
oh im srry never mind i didnt see the seconds ur answer is correct
im 100% sure its right
I agree - your work and answer is correct
do ya take entrepuener class
what do you mean??
other than physics do u take entrepuener class
its like a business class
no!! never!!! why??
haha i kinda needed help
ok lol but if u dont knw it u dont have to help k
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
(finance) the net sales minus the cost of goods and services sold.
i gave u a medal right
cool so did u graduate? or u still in high skool