A ball is launched horizontally at 5 m s-1 from the end of a table. The ball is in flight for 0.4 s

before it lands on the floor. The ball is now launched from the end of the same table with a

horizontal velocity 10 m s-1.

What is the new time of flight of the ball?

A 0.2 s

B 0.4 s

C 0.5 s

D 0.8 s

The answer is B here, but im confused how do i prove that time of flight is independent of horizontal velocity?

before it lands on the floor. The ball is now launched from the end of the same table with a

horizontal velocity 10 m s-1.

What is the new time of flight of the ball?

A 0.2 s

B 0.4 s

C 0.5 s

D 0.8 s

The answer is B here, but im confused how do i prove that time of flight is independent of horizontal velocity?

Hi, using my extensive physics knowledge, I think it depends on the country. This is not an exam style question, not enough information

Original post by efsfsf\\eeeeeee

A ball is launched horizontally at 5 m s-1 from the end of a table. The ball is in flight for 0.4 s

before it lands on the floor. The ball is now launched from the end of the same table with a

horizontal velocity 10 m s-1.

What is the new time of flight of the ball?

A 0.2 s

B 0.4 s

C 0.5 s

D 0.8 s

The answer is B here, but im confused how do i prove that time of flight is independent of horizontal velocity?

before it lands on the floor. The ball is now launched from the end of the same table with a

horizontal velocity 10 m s-1.

What is the new time of flight of the ball?

A 0.2 s

B 0.4 s

C 0.5 s

D 0.8 s

The answer is B here, but im confused how do i prove that time of flight is independent of horizontal velocity?

The time of flight is determined by vertical motion so the vertical velocity and vertical acceleration which determines the time taken to return to zero (vertical) height. Its completely independent of horizontal velocity.

In both cases, the initial vertical velocity is zero, and the vertical acceleration is acceleration due to gravity, which is also constant, since the height of the table is also constant, then the time of flight is constant, according to the relation h = 1/2g(t^2)

Original post by mqb2766

The time of flight is determined by vertical motion so the vertical velocity and vertical acceleration which determines the time taken to return to zero (vertical) height. Its completely independent of horizontal velocity.

What about s = ut though, derived from s=ut+1/2at^2 where there is no acc horizontally, doesnt that show that initial speed and time are related if horizontal displacement is constant?

Original post by emmanuel afolabi

In both cases, the initial vertical velocity is zero, and the vertical acceleration is acceleration due to gravity, which is also constant, since the height of the table is also constant, then the time of flight is constant, according to the relation h = 1/2g(t^2)

what about s =ut for horizontal displacement?

also does it matter that initial vertical velocity is 0 or does it matter that its constant?

(edited 2 months ago)

Original post by efsfsf\\eeeeeee

What about s = ut though, derived from s=ut+1/2at^2 where there is no acc horizontally, doesnt that show that initial speed and time are related if horizontal displacement is constant?

Vertical and horiziontal motion are completely independent. Gravity (vertical acceleration) does not affect horizontal motion and here the horizontal initial velocity has no effect on the vertical motion.

Proof by video

https://www.youtube.com/watch?v=beyX4jO_NUE

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