Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ A ball is thrown horizontally from the top of a building 19 m high. The ball strikes the ground at a point 98.2 m from the base of the building. The acceleration of gravity is 9.8 m/s\(^2\).

Find the time the ball is in motion.

To find the time the ball is in motion, we need to consider the vertical motion of the ball as it falls from the top of the building to the ground. Using the equations for vertical and horizontal motion, we can calculate the time.

To find the time the ball is in motion, we need to consider the vertical motion of the ball as it falls from the top of the building to the ground.

First, we can use the equation y = ut + 1/2gt^2, where y is the vertical displacement, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.

In this case, the initial vertical velocity is 0 m/s since the ball is thrown horizontally.

Since the ball lands 100.0 m from the base of the building, we can use the equation x = ut, where x is the horizontal displacement and u is the initial horizontal velocity. We can rearrange this equation to solve for time, t = x/u.

Using the given information, we can calculate the time the ball is in motion.