Question

A ball is thrown vertically upwards 40 metres per second : calculate the greatest height attained and its velocity after 3 seconds​

Answer 1

Answer:

Given Initial velocity of ball, u=49 m/s

Given Initial velocity of ball, u=49 m/sLet the maximum height reached and time taken to reach that height be H and t respectively.

Given Initial velocity of ball, u=49 m/sLet the maximum height reached and time taken to reach that height be H and t respectively.Assumption: g=9.8 m/s2 holds true (maximum height reached is small compared to the radius of earth)

Given Initial velocity of ball, u=49 m/sLet the maximum height reached and time taken to reach that height be H and t respectively.Assumption: g=9.8 m/s2 holds true (maximum height reached is small compared to the radius of earth)Velocity of the ball at maximum height is zero, v=0

Given Initial velocity of ball, u=49 m/sLet the maximum height reached and time taken to reach that height be H and t respectively.Assumption: g=9.8 m/s2 holds true (maximum height reached is small compared to the radius of earth)Velocity of the ball at maximum height is zero, v=0v2−u2=2aH                          

Given Initial velocity of ball, u=49 m/sLet the maximum height reached and time taken to reach that height be H and t respectively.Assumption: g=9.8 m/s2 holds true (maximum height reached is small compared to the radius of earth)Velocity of the ball at maximum height is zero, v=0v2−u2=2aH                          0−(49)2=2×(−9.8)×H         

Given Initial velocity of ball, u=49 m/sLet the maximum height reached and time taken to reach that height be H and t respectively.Assumption: g=9.8 m/s2 holds true (maximum height reached is small compared to the radius of earth)Velocity of the ball at maximum height is zero, v=0v2−u2=2aH                          0−(49)2=2×(−9.8)×H         ⟹H=122.5 m               

Given Initial velocity of ball, u=49 m/sLet the maximum height reached and time taken to reach that height be H and t respectively.Assumption: g=9.8 m/s2 holds true (maximum height reached is small compared to the radius of earth)Velocity of the ball at maximum height is zero, v=0v2−u2=2aH                          0−(49)2=2×(−9.8)×H         ⟹H=122.5 m               v=u+at

Given Initial velocity of ball, u=49 m/sLet the maximum height reached and time taken to reach that height be H and t respectively.Assumption: g=9.8 m/s2 holds true (maximum height reached is small compared to the radius of earth)Velocity of the ball at maximum height is zero, v=0v2−u2=2aH                          0−(49)2=2×(−9.8)×H         ⟹H=122.5 m               v=u+at0=49−9.8t            

Given Initial velocity of ball, u=49 m/sLet the maximum height reached and time taken to reach that height be H and t respectively.Assumption: g=9.8 m/s2 holds true (maximum height reached is small compared to the radius of earth)Velocity of the ball at maximum height is zero, v=0v2−u2=2aH                          0−(49)2=2×(−9.8)×H         ⟹H=122.5 m               v=u+at0=49−9.8t            ⟹t=5 s

Given Initial velocity of ball, u=49 m/sLet the maximum height reached and time taken to reach that height be H and t respectively.Assumption: g=9.8 m/s2 holds true (maximum height reached is small compared to the radius of earth)Velocity of the ball at maximum height is zero, v=0v2−u2=2aH                          0−(49)2=2×(−9.8)×H         ⟹H=122.5 m               v=u+at0=49−9.8t            ⟹t=5 s∴ Total time taken by ball to return to the surface, T=2t=10 s 

Explanation:

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Answer 2

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