Question

You throw a ball horizontally off of a building with a horizontal velocity of 5 m/s. What is the balls horizontal velocity 8 seconds later? How high should the building be if you want the ball to fall for a total of 12 seconds?

Answer 1

The horizontal velocity of the ball remains at 5 m/s after 8 seconds as there's no horizontal acceleration. To fall for 12 seconds, the building needs to be 705.6 meters high, calculated using the displacement formula for free fall under gravity.

Step-by-step explanation:

The question involves projectile motion, specifically the horizontal motion of a projectile when the only force acting on it is gravity. When a ball is thrown horizontally with an initial velocity and no forces other than gravity act upon it, the horizontal velocity remains constant since there's no acceleration in that direction. Therefore, the ball's horizontal velocity 8 seconds later remains at 5 m/s.

For the ball to fall for a total of 12 seconds, one must determine how high the building should be. The formula to calculate the height when the only force is gravity is h = 1/2 * g * t^2, where g is the acceleration due to gravity (9.8 m/s^2) and t is the time in seconds. Plugging in the values:

h = 1/2 * 9.8 m/s^2 * (12 s)^2

h = 1/2 * 9.8 m/s^2 * 144 s^2

h = 1/2 * 9.8 m/s^2 * 144

h = 705.6 meters

The building should be 705.6 meters high for the ball to be in free fall for 12 seconds.