A ball is kicked with an initial velocity of 16.1 m/s in the horizontal direction and 14.4 m/s in the vertical direction. What maximum height is attained by the ball?
The horizontal speed does not matter.
As you know, the max height is
v^2/2g
where v is the vertical speed. So, plug in your numbers.
To find the maximum height attained by the ball, we need to determine the vertical component of its motion. We can use the equations of motion to find the time it takes for the ball to reach its maximum height.
First, let's find the time it takes for the ball to reach its highest point. We can use the equation: vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is time.
In this case, the initial vertical velocity (vi) is 14.4 m/s, the final vertical velocity (vf) is 0 m/s (at the maximum height, the ball momentarily stops before falling), and the acceleration (a) is the acceleration due to gravity, which is approximately -9.8 m/s^2 (negative because it acts in the opposite direction to the initial velocity).
0 = 14.4 - 9.8t
Now, let's solve for t:
9.8t = 14.4
t = 14.4 / 9.8
t ≈ 1.469 s
Next, we can use the equation to find the maximum height (hmax) attained by the ball:
hmax = vi * t + (1/2) * a * t^2
Substituting the values:
hmax = 14.4 * 1.469 + (1/2) * (-9.8) * (1.469)^2
Calculating the expression:
hmax ≈ 21.12 m
Therefore, the maximum height attained by the ball is approximately 21.12 meters.