Respuesta :
Answer:
h = 9.57 seconds
Explanation:
It is given that,
Initial speed of Kalea, u = 13.7 m/s
At maximum height, v = 0
Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :
[tex]v=u-gt[/tex]
[tex]u=gt[/tex]
[tex]t=\dfrac{u}{g}[/tex]
[tex]t=\dfrac{13.7}{9.8}[/tex]
t = 1.39 s
Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :
[tex]h=ut+\dfrac{1}{2}at^2[/tex]
Here, a = -g
[tex]h=ut-\dfrac{1}{2}gt^2[/tex]
[tex]h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2[/tex]
h = 9.57 meters
So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.
The ball thrown by Kalea is in an upward direction so the maximum height obtained by the ball is in 1.39 seconds.
The maximum height obtained by the ball is 9.57 meters.
What is the height?
A height can be defined as the maximum vertical distance covered by an object.
Given that the initial velocity is 13.7 m/s and g is 9.80.
Kalea throws a baseball directly upward, then at the highest point, the final velocity is zero.
Let t is the time taken by the ball to reach its maximum point then,
[tex]v=u-gt[/tex]
[tex]0=13.7 - 9.8\times t[/tex]
[tex]t = 1.39 \;\rm s[/tex]
Let h is the height reached by the ball above its release point,
[tex]h = ut +\dfrac {1}{2}at^2[/tex]
Here, a = -g
[tex]h = ut-\dfrac{1}{2}gt^2[/tex]
[tex]h = 13.7\times1.39-\dfrac{1}{2}\times9.8\times 1.39^2[/tex]
[tex]h = 9.57 \;\rm m[/tex]
Hence we can conclude that the height obtained by the ball is 9.57 meters.
To know more about the velocity and distance, follow the link given below.
https://brainly.com/question/12626613.
