I need full solution to this 2 questions then I will mark as brainliest and follow u
Answers
Answer:
First Question :
When ball is falling down from height d and reaches at height h , then velocity acquired by the ball is
When ball is falling down from height d and reaches at height h , then velocity acquired by the ball isv2= 2g(d-h) —(i)
When ball is falling down from height d and reaches at height h , then velocity acquired by the ball isv2= 2g(d-h) —(i)This is downward velocity and is taken as negative .
When ball is falling down from height d and reaches at height h , then velocity acquired by the ball isv2= 2g(d-h) —(i)This is downward velocity and is taken as negative .When h = d , v = 0
When ball is falling down from height d and reaches at height h , then velocity acquired by the ball isv2= 2g(d-h) —(i)This is downward velocity and is taken as negative .When h = d , v = 0When h =0 , v2 = (2 g d)
When ball is falling down from height d and reaches at height h , then velocity acquired by the ball isv2= 2g(d-h) —(i)This is downward velocity and is taken as negative .When h = d , v = 0When h =0 , v2 = (2 g d)Hence from eqn(i) , the variation of v & h is parabolic below h axis
ANSWER TO SECOND QUESTION :
As we know that acceleration is the rate of change of the velocity.
As we know that acceleration is the rate of change of the velocity.a =acceleration m/s²
As we know that acceleration is the rate of change of the velocity.a =acceleration m/s²v= velocity m/s
As we know that acceleration is the rate of change of the velocity.a =acceleration m/s²v= velocity m/sdv= a.dt
As we know that acceleration is the rate of change of the velocity.a =acceleration m/s²v= velocity m/sdv= a.dtv= ∫a .dt
As we know that acceleration is the rate of change of the velocity.a =acceleration m/s²v= velocity m/sdv= a.dtv= ∫a .dtWe can say that area of the acceleration graph is equal to the velocity.
As we know that acceleration is the rate of change of the velocity.a =acceleration m/s²v= velocity m/sdv= a.dtv= ∫a .dtWe can say that area of the acceleration graph is equal to the velocity.After the point B acceleration of train become negative it means that at point B train will have maximum velocity.So the area of graph between O and B
As we know that acceleration is the rate of change of the velocity.a =acceleration m/s²v= velocity m/sdv= a.dtv= ∫a .dtWe can say that area of the acceleration graph is equal to the velocity.After the point B acceleration of train become negative it means that at point B train will have maximum velocity.So the area of graph between O and BA=1/2 x 5 x 4 + 5 x (8-4) m/s
As we know that acceleration is the rate of change of the velocity.a =acceleration m/s²v= velocity m/sdv= a.dtv= ∫a .dtWe can say that area of the acceleration graph is equal to the velocity.After the point B acceleration of train become negative it means that at point B train will have maximum velocity.So the area of graph between O and BA=1/2 x 5 x 4 + 5 x (8-4) m/sA= 10 + 20 m/s
As we know that acceleration is the rate of change of the velocity.a =acceleration m/s²v= velocity m/sdv= a.dtv= ∫a .dtWe can say that area of the acceleration graph is equal to the velocity.After the point B acceleration of train become negative it means that at point B train will have maximum velocity.So the area of graph between O and BA=1/2 x 5 x 4 + 5 x (8-4) m/sA= 10 + 20 m/sA= 30 m/s
As we know that acceleration is the rate of change of the velocity.a =acceleration m/s²v= velocity m/sdv= a.dtv= ∫a .dtWe can say that area of the acceleration graph is equal to the velocity.After the point B acceleration of train become negative it means that at point B train will have maximum velocity.So the area of graph between O and BA=1/2 x 5 x 4 + 5 x (8-4) m/sA= 10 + 20 m/sA= 30 m/sSo the maximum velocity is 30 m/s.