Question

A projectile is fired vertically into the air. Its motion is described by h = -16t^2 + 320t where h is the projectile's height in feet after t seconds.

A) During what time intervals will the height of the projectile be below 1024 feet?
B) During what time intervals will the height of the projectile be above 1024 feet?
C) When will the projectile land back on the ground?

Answer 1

The height of the projectile can be determined using the equation h = -16t^2 + 320t. The time intervals when the height is below 1024 feet are t < 2 and t > 20. The time interval when the height is above 1024 feet is 2 < t < 20. The projectile lands back on the ground at t = 0 and t = 20.

Step-by-step explanation:

The height of a projectile fired vertically into the air can be determined using the equation h = -16t^2 + 320t, where h is the height in feet and t is the time in seconds.

A) To find the time intervals when the height of the projectile is below 1024 feet, we set the equation h = -16t^2 + 320t < 1024 and solve for t. This gives us two time intervals: t < 2 and t > 20.

B) To find the time intervals when the height of the projectile is above 1024 feet, we set the equation h = -16t^2 + 320t > 1024 and solve for t. This gives us the time interval 2 < t < 20.

C) To find when the projectile lands back on the ground, we set the equation h = -16t^2 + 320t = 0 and solve for t. This gives us t = 0 and t = 20.