You throw a ball horizontally off of a building with a horizontal velocity of 10 m/s. What is the ball’s horizontal velocity 5 seconds later? How high should the building be if you want the ball to fall for a total of 10 seconds?
Vh = 10m/s and is constant.
h = Vo*t + 0.5gt^2,
h = 0 + 4.9(10)^2 = 490m.
I response my question
To find the ball's horizontal velocity 5 seconds later, we first need to understand that the horizontal velocity of a projectile remains constant throughout its motion (assuming no horizontal forces acting on it).
In this case, the ball's horizontal velocity is given as 10 m/s. So, regardless of the time elapsed, the horizontal velocity will remain 10 m/s.
Therefore, the ball's horizontal velocity 5 seconds later will still be 10 m/s.
Now, let's find out how high the building should be to make the ball fall for a total of 10 seconds.
The time it takes for an object in free fall to fall from a certain height can be calculated using the equation:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of fall.
In this case, we want the ball to fall for a total of 10 seconds, so we can rearrange the equation to solve for h:
h = (1/2) * g * t^2 = (1/2) * 9.8 * 10^2 = 490 meters
Therefore, the building should be 490 meters high for the ball to fall for a total of 10 seconds.