Question

Andy shoots on goal in lacrosse, and instead of "bar down," the ball goes straight up into the air with an initial velocity of 17m/s. how high does the ball go? how long is it in the air?

Answer 1

The ball reaches a maximum height of approximately 14.22 meters and remains in the air for approximately 1.735 seconds.

Step-by-step explanation:

To determine the maximum height and the time the ball remains in the air, we can analyze the vertical motion of the ball. We can use the kinematic equations of motion to solve for these values.

1. Maximum height: The ball reaches its maximum height when its vertical velocity becomes zero. We can use the equation final velocity squared = initial velocity squared + 2 * acceleration * displacement to find the displacement. In this case, the final velocity is 0 m/s, the initial velocity is 17 m/s, and the acceleration is -9.8 m/s^2 (due to gravity). Solving for displacement, we get:

vf2 = vi2 + 2 * a * d

0 = (17)^2 + 2 * (-9.8) * d

d = ((17)^2) / (2 * (-9.8))

d ≈ 14.22 m

Therefore, the maximum height the ball reaches is approximately 14.22 meters.

1. Time in the air: The time the ball remains in the air can be determined using the equation final velocity = initial velocity + (acceleration * time). Since the final velocity is 0 m/s and the initial velocity is 17 m/s, we can solve for time:

vf = vi + a * t

0 = 17 + (-9.8) * t

t ≈ 1.735 s

Therefore, the ball remains in the air for approximately 1.735 seconds.