Question

you throw a ball at an initial velocity of 22m/s at an angle of 40 degrees above the horizontal. what is the max height of the ball and how far will it travel in the horizontal direction?

Answer 1

To find the maximum height of the ball, you need to analyze the vertical motion. The maximum height can be found using the equation max_height = vertical_velocity^2 / (2 * g), where vertical_velocity = initial_velocity * sin(angle). The horizontal distance traveled by the ball can be found using the equation horizontal_distance = horizontal_velocity * time, where horizontal_velocity = initial_velocity * cos(angle) and time = 2 * max_height / g.

Step-by-step explanation:

To find the maximum height of the ball, we need to analyze the vertical motion.

First, we can decompose the initial velocity into its vertical and horizontal components using trigonometry.

The vertical component is given by vertical_velocity = initial_velocity * sin(angle).

In this case, the vertical component is 22 m/s * sin(40).

Next, we can use the equation max_height = vertical_velocity^2 / (2 * g), where g is the acceleration due to gravity. Plugging in the values, we get max_height = (22 * sin(40))^2 / (2 * 9.8).

To find the horizontal distance traveled by the ball, we can use the equation horizontal_distance = horizontal_velocity * time. The horizontal velocity is given by horizontal_velocity = initial_velocity * cos(angle).

In this case, the horizontal velocity is 22 m/s * cos(40).

The time can be found using the equation time = 2 * max_height / g.

Plugging in the values, we get horizontal_distance = (22 * cos(40)) * (2 * (22 * sin(40))^2 / (2 * 9.8) / 9.8).