Verified answer

Answer:

• 2 second is the required answer

Explanation:

According to the Question

It is given that,

• Initial velocity ,u = 72km/h
• Final velocity ,v = 0m/s
• Acceleration due to gravity ,g = 10ms⁻²

we have to calculate the time taken by ball to reach at maximum height .

Firstly we convert the unit here

↠ u = 72×5/18 m/s

↠ u = 20m/s

Now, using Kinematic Equation

• v = u + gt

On substituting the value we get

↠ 0 = 20 + (-10) × t

↠ -20 = -10 × t

↠ -20/-10 = t

↠ t = 20/10

↠ t = 2s

• Hence, the time taken by the ball to reach the maximum height is 2 second.

To find :-

The time taken to reach at maximum height :-

We know that the following :-

Initial Velocity = > 72 km/h

Final velocity = > 0

Acceleration due to gravity = > 9.8 or 10

Convert 72 km/h in m/s

= > 20 m/s

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Solution :-

= > Now we have to find the time at maximum height :-

We use 1st eq of motion :- v = u + at

v = u + at

0 = 20 + (-10)t (We take gravitation in minus because the ball is thrown in upward direction)

-20 = -10 × t

[tex] \frac{ - 20}{ - 10} = t [/tex]

2 sec = time

Therefore, The time taken to reach at maximum height is 2 sec.

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