see u=10 m/s and at maximum height v=0 m/s so time required to reach maximum height is t= u/g i.e. 10 /10 = 1 s.now maximum height h= 1/2gt×t =1/2×10×1×1=5 m/s

Analysis:

To find the maximum height attained by the ball, we can use the kinematic equation for vertical motion. The equation we can use is:

v² = u² + 2as

where:
v = final velocity (9 m/s)
u = initial velocity (10 m/s)
a = acceleration due to gravity (-9.8 m/s², considering downward direction as negative)
s = displacement (maximum height attained)

Calculation:

Using the given values, we can rearrange the equation to solve for s:

s = (v² - u²) / (2a)

Substituting the values:

s = (9² - 10²) / (2 * -9.8)
s = (81 - 100) / (-19.6)
s = -19 / -19.6
s ≈ 0.969 m

Explanation:

The negative sign in the equation signifies that the displacement is in the opposite direction of the acceleration due to gravity. In this case, the ball is thrown upwards, so the displacement is positive.

The maximum height attained by the ball is approximately 0.969 m. This means that the ball reaches a height of 0.969 m above its initial position before falling back to the ground.

Summary:

- The ball is thrown vertically upwards with an initial velocity of 10 m/s.
- It returns to the ground with a final velocity of 9 m/s.
- Using the kinematic equation for vertical motion, we can find the maximum height attained by the ball.
- The maximum height attained is approximately 0.969 m.