You throw a ball horizontally off of a building with a horizontal velocity of 5 m/s. What's the ball horizontal velocity 8 seconds later? How high should the building be if you want the ball to fall for a total of 12 seconds?
The horizontal velocity is constant.
h=1/2 g 12^2= 1/2 *9.8*144 meters.
aregergerger
To answer the first question, we can assume that there is no horizontal acceleration acting on the ball, meaning its horizontal velocity remains constant. So, if the ball is thrown horizontally with a velocity of 5 m/s, the horizontal velocity of the ball will still be 5 m/s after 8 seconds. This is because no force is acting on the ball to change its horizontal velocity.
Now, let's move on to the second question. To determine the height of the building, we need to consider the motion of the ball in the vertical direction. Assuming the ball is subject to only the force of gravity, its vertical motion can be described using the equation:
y = v0 * t + (1/2) * g * t^2
where:
- y is the vertical displacement or height of the ball
- v0 is the initial vertical velocity of the ball (which is zero in this case since it is only thrown horizontally)
- t is the time elapsed
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
We want the ball to fall for a total of 12 seconds, so if we substitute t = 12 into the equation, the equation becomes:
y = 0 + (1/2) * 9.8 * (12^2)
Simplifying the equation:
y = 0 + 0.5 * 9.8 * 144
y = 0 + 70.56
y = 70.56 meters
Therefore, the building should be approximately 70.56 meters high in order for the ball to fall for a total of 12 seconds.