Final answer:
After 1 second, the ball is approximately 4.9 feet in the air. After 2 seconds, the ball is approximately 10 feet in the air.
Step-by-step explanation:
To answer the question, we need to consider the motion of the ball. If the ball is thrown straight up and falls back down, we can use the equation of motion to calculate its height at different times.
After 1 second, the ball is approximately 4.9 feet in the air. After 2 seconds, the ball reaches its highest point and then starts to fall back down, so it is approximately 10 feet in the air.After 1 second, the ball is typically a certain height above the ground, and this value depends on the initial conditions of the ball's motion. Assuming the ball is thrown vertically upward, its height at any given time can be described by the physics formula:
ℎ()=0−122h(t)=v 0 t− 21 gt 2 ,
where ℎ()h(t) is the height, 0v0 is the initial velocity, g is the acceleration due to gravity (approximately 32 feet per second squared). After 1 second (=1t=1), you can substitute this value into the formula to find the height. Similarly, after 2 seconds (=2t=2), you can calculate the new height. The values will depend on the specific initial conditions of the ball's motion, such as the initial velocity when thrown. The formula helps determine the ball's position at different time intervals in its upward trajectory.