A ball is thrown upward.What is its maximum height? Its initial vertical speed is 11.5 m/s and the acceleration

of gravity is 9.8 m/s2 . Neglect air resistance.
Answer in units of m

20.856 20.856m

To determine the maximum height reached by the ball, we can use the kinematic equation for vertical position:

y = y0 + v0t - (1/2)gt^2

where:
- y is the vertical position at a given time, t
- y0 is the initial vertical position (which we can assume to be zero since the ball starts from the ground)
- v0 is the initial vertical speed (11.5 m/s)
- g is the acceleration due to gravity (-9.8 m/s^2, considering upward motion)
- t is the time elapsed

At the maximum height, the final vertical velocity will be zero since the ball momentarily stops before falling back down. Therefore, we can set v = 0 in the kinematic equation:

0 = v0 - gt

Solving for t:

t = v0 / g
= 11.5 m/s / 9.8 m/s^2
≈ 1.173 seconds

Now, substitute this value of t back into the original kinematic equation to find the maximum height:

y = y0 + v0t - (1/2)gt^2
= 0 + (11.5 m/s)(1.173 s) - (1/2)(9.8 m/s^2)(1.173 s)^2
≈ 6.03 meters

Therefore, the maximum height reached by the ball is approximately 6.03 meters.