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She then throws a ball upwards with an initial velocity of 28 m s¹ at 45° to the horizontal. Her hand was 1.6 m above the ground. A short time later the ball was caught by another player. When it was caught the ball was again at a height of 1.6 m.
Calculate how long the ball was in the air?​

Verified Answer :

nishabiju31

Answer:

The solutions are

=

0

t=0 and

=

14

4.9

≈

2.86

t=

4.9

14

≈2.86 seconds. Since time can't be negative, the ball was in the air for approximately 2.86 seconds.

Explanation:

To find out how long the ball was in the air, you can use the vertical motion equation:

ℎ

=

ℎ

0

+

0

−

1

2

2

h=h

0

+v

0y

t−

2

1

gt Where:

ℎ

h is the final height (1.6 m when caught),

ℎ

0

h

0

is the initial height (1.6 m when thrown),

0

v

0y

is the vertical component of the initial velocity (since it's thrown at an angle,

0

=

0

sin

⁡

(

)

v

0y

=v

0

sin(θ)),

g is the acceleration due to gravity (approximately

9.8

 

m/s

2

9.8m/s

2

),

t is the time in seconds.Given:

ℎ

=

1.6

 

m

h=1.6m,

ℎ

0

=

1.6

 

m

h

0

=1.6m,

0

=

28

 

m/s

v

0

=28m/s,

=

4

5

∘

θ=45

∘

,

=

9.8

 

m/s

2

g=9.8m/s

2First, find

0

v

0y

:

0

=

0

sin

⁡

(

)

v

0y

=v

0

sin(θ)

0

=

28

 

m/s

×

sin

⁡

(

4

5

∘

)

v

0y

=28m/s×sin(45

∘

)

0

=

28

 

m/s

×

2

2

v

0y

=28m/s×

2

2

0

=

14

 

m/s

v

0y

=14m/s

Now, plug in the values to solve for

t:

1.6

=

1.6

+

(

14

)

−

1

2

(

9.8

)

2

1.6=1.6+(14)t−

2

1

(9.8)t

2

This simplifies to:

0

=

14

−

4.9

2

0=14t−4.9t

2

Solve this quadratic equation for

t.

2

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