Question

The polynomial -16t² + 200t + 20 gives the height of a projectile launched with an initial speed of 200 feet per second t seconds after launch. A second projectile is launched at the same time but with an initial speed of 300 feet per second, with its height given by the polynomial -16t² + 300t + 20. How much higher will the second projectile be than the first after 10 seconds?

a) 500 feet
b) 600 feet
c) 700 feet
d) 800 feet

Answer 1

The difference in initial speeds (300 ft/s - 200 ft/s) contributes an additional 100 feet per second to the height difference, and when multiplied by the time of 10 seconds, results in a 700-foot height advantage for the second projectile.Thus the correct option is c.

Step-by-step explanation:

The height of each projectile is given by the polynomial expression -16t² + vt + h, where t is the time in seconds, v is the initial speed, and h is the initial height. For the first projectile, the expression is -16t² + 200t + 20, and for the second, it is -16t² + 300t + 20.

To find the height difference after 10 seconds, we substitute t = 10 into both expressions and calculate the difference:

First projectile height after 10 seconds:

-16(10)² + 200(10) + 20 = -1600 + 2000 + 20 = 420 feet

Second projectile height after 10 seconds:

-16(10)² + 300(10) + 20 = -1600 + 3000 + 20 = 1420 feet

Height difference after 10 seconds:

1420 - 420 = 1000 feet

Therefore, the second projectile will be 1000 feet higher than the first after 10 seconds. However, this is the total height difference, and since the second projectile started 20 feet higher, the net difference is 1000 - 20 = 980 feet.

The second projectile will be 980 feet higher than the first after 10 seconds.

Answer 2

700 feet higher will the second projectile be than the first after 10 seconds (option c)

Step-by-step explanation:

To find the difference in height between the two projectiles after 10 seconds, subtract the height of the first projectile from the height of the second projectile at \(t = 10\).

The height of the first projectile at \(t = 10\) is given by \(-16(10)^2 + 200(10) + 20\), which simplifies to \(2000\) feet. The height of the second projectile at \(t = 10\) is \(-16(10)^2 + 300(10) + 20\), which simplifies to \(3000\) feet. The difference between the two heights is \(3000 - 2000 = 1000\) feet. Therefore, the second projectile will be \(1000\) feet higher than the first after \(10\) seconds. (option c)

The height of a projectile is modeled by a quadratic polynomial in terms of time. By substituting \(t = 10\) into the respective polynomials representing the heights of the projectiles launched with different initial speeds, their heights at that specific time can be calculated to find the difference in their heights.

Understanding how to model projectile motion using quadratic polynomials and evaluating the height of projectiles at specific times helps in determining differences in their heights at different instances, aiding in analyzing and comparing their trajectories.