Question

You should compile a list of values, including known and unknown quantities. Identify the following terms as knowns or unknowns. A ball thrown horizontally with speed vo 23.0 m/s travels a horizontal distance of d 55.0 m before hitting the ground. From what height h was the ball thrown?

Answer 1

To find the height from which a ball is thrown horizontally, we identify the initial speed, horizontal distance, and acceleration due to gravity as knowns, and calculate the time in the air to determine the unknown height using kinematic equations.

Step-by-step explanation:

Identifying Knowns and Unknowns in Projectile Motion

To solve for the unknown height h from which a ball is thrown horizontally, we identify the knowns and unknowns. The knowns include the initial horizontal speed vo = 23.0 m/s, the horizontal distance d = 55.0 m, and the acceleration due to gravity a = -9.8 m/s² (since it's directed downwards). The time t the ball is in the air is an unknown that we can calculate using the horizontal motion because there is no horizontal acceleration. After finding t, we can use it to calculate the height h, which is the other unknown in this problem. The vertical motion can be examined in isolation to find the height using the kinematic equations for motion.

Steps for Solving

1. Use the horizontal motion equation d = vo * t to find the time in the air (t).
2. Apply the vertical motion equation h = 0.5 * a * t² to find the height (h).

Answer 2

The ball was thrown from a height of 66.775 meters.

Explanation:

The key values in this problem are:

Knowns: Horizontal speed (vo) = 23.0 m/s, horizontal distance (d) = 55.0 m, acceleration due to gravity (g) = 9.8 m/s²

Unknown: Initial height (h)

To find the initial height (h), we can use the horizontal motion equation:
`\(d = (1)/(2)gt^2\)`, where (t) is the time of flight. Since the ball is thrown horizontally, the vertical motion is governed by
`\(h = (1)/(2)gt^2\)`. The time of flight can be determined from the horizontal motion using
`\(t = (d)/(v_0)\)`.

Substituting this into the vertical motion equation gives
`\(h = (1)/(2)g\left((d)/(v_0)\right)^2\)`. Plugging in the known values, we find
`\(h = (1)/(2)(9.8)\left((55)/(23)\right)^2 = 66.775\)` meters.