Question

the time taken to return to its starting point.​

Answer 1

Answer:

Answer:Good morning friend.

Answer:Good morning friend. here is your answer

Answer:Good morning friend. here is your answerLet the maximum height reached be H.

Answer:Good morning friend. here is your answerLet the maximum height reached be H.At the maximum height, final velocity of ball is zero i.e. v=0

Answer:Good morning friend. here is your answerLet the maximum height reached be H.At the maximum height, final velocity of ball is zero i.e. v=0Initial velocity u=+20 m/s (Considering upward direction to be positive)

Answer:Good morning friend. here is your answerLet the maximum height reached be H.At the maximum height, final velocity of ball is zero i.e. v=0Initial velocity u=+20 m/s (Considering upward direction to be positive)Acceleration due to gravity g=−9.8 m/s

Answer:Good morning friend. here is your answerLet the maximum height reached be H.At the maximum height, final velocity of ball is zero i.e. v=0Initial velocity u=+20 m/s (Considering upward direction to be positive)Acceleration due to gravity g=−9.8 m/s 2

Answer:Good morning friend. here is your answerLet the maximum height reached be H.At the maximum height, final velocity of ball is zero i.e. v=0Initial velocity u=+20 m/s (Considering upward direction to be positive)Acceleration due to gravity g=−9.8 m/s 2

Answer:Good morning friend. here is your answerLet the maximum height reached be H.At the maximum height, final velocity of ball is zero i.e. v=0Initial velocity u=+20 m/s (Considering upward direction to be positive)Acceleration due to gravity g=−9.8 m/s 2 Using v=u+gt

Answer:Good morning friend. here is your answerLet the maximum height reached be H.At the maximum height, final velocity of ball is zero i.e. v=0Initial velocity u=+20 m/s (Considering upward direction to be positive)Acceleration due to gravity g=−9.8 m/s 2 Using v=u+gtOr 0=20−9.8t

Answer:Good morning friend. here is your answerLet the maximum height reached be H.At the maximum height, final velocity of ball is zero i.e. v=0Initial velocity u=+20 m/s (Considering upward direction to be positive)Acceleration due to gravity g=−9.8 m/s 2 Using v=u+gtOr 0=20−9.8t⟹ t≈2 s

Answer:Good morning friend. here is your answerLet the maximum height reached be H.At the maximum height, final velocity of ball is zero i.e. v=0Initial velocity u=+20 m/s (Considering upward direction to be positive)Acceleration due to gravity g=−9.8 m/s 2 Using v=u+gtOr 0=20−9.8t⟹ t≈2 sWe have found the time of ascending of ball to be 2 s.

Answer:Good morning friend. here is your answerLet the maximum height reached be H.At the maximum height, final velocity of ball is zero i.e. v=0Initial velocity u=+20 m/s (Considering upward direction to be positive)Acceleration due to gravity g=−9.8 m/s 2 Using v=u+gtOr 0=20−9.8t⟹ t≈2 sWe have found the time of ascending of ball to be 2 s.Time of descending of ball is also 2 s

Answer:Good morning friend. here is your answerLet the maximum height reached be H.At the maximum height, final velocity of ball is zero i.e. v=0Initial velocity u=+20 m/s (Considering upward direction to be positive)Acceleration due to gravity g=−9.8 m/s 2 Using v=u+gtOr 0=20−9.8t⟹ t≈2 sWe have found the time of ascending of ball to be 2 s.Time of descending of ball is also 2 sThus total time taken to reach the ground T=2+2=4 s

Answer 2

Answer:

The ball will spend 1/2 the total time going up and the same amount of time coming down.

This is useful to know, because at maximum height the ball has a speed of zero.

Initial speed (u) = 30 m/s end speed (v) = 0

Acceleration (a) = -9.81 m/s^2 time (t) = unknown

Now we can substitute these values into a well known equation for kinematics, under constant acceleration: v = u + at

We find the time (t) taken to get the top and then double it to get the total time of flight.

0 = 30 + (- 9.81*t)

t = 30/9.81

t = 3.058 seconds

Double it for total time in the air = 6.12 seconds

(2 decimal places)

Explanation:

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