Question

PLEASE HELP

During a game’s intermission, a team mascot launches tightly rolled tee shirts into the stands. The height h(t), in feet, of an airborne tee shirt t seconds after being launched can be approximated by: ℎ() = −8 2 + 35 + 5

a. How high is the tee shirt 1 second after being thrown?

b. After peaking, a rolled-up tee shirt is caught by a fan 40 ft above the ground level. For how long was the tee shirt in the air?

c. The team mascot throws another tee shirt and misses the stands. How long was the teeshirt in the air when it hit the ground?​

Answer 1

To answer the questions, we'll use the given function h(t) = -8t^2 + 35t + 5.

a. To find the height of the tee shirt 1 second after being thrown, we can substitute t = 1 into the function:

h(1) = -8(1)^2 + 35(1) + 5

h(1) = -8 + 35 + 5

h(1) = 32 feet

So, the tee shirt is approximately 32 feet high 1 second after being thrown.

b. To find how long the tee shirt was in the air, we need to find the time at which h(t) = 40. We solve the quadratic equation:

-8t^2 + 35t + 5 = 40

-8t^2 + 35t - 35 = 0

Using the quadratic formula, we get t = 4.5 seconds or t = -1.125 seconds. Since time can't be negative, the tee shirt was in the air for 4.5 seconds before it was caught.

c. When the tee shirt hits the ground, its height is 0. To find out how long the tee shirt was in the air before hitting the ground, we set h(t) = 0 and solve for t:

-8t^2 + 35t + 5 = 0

Using the quadratic formula, we get t ≈ 4.60 seconds or t ≈ 0.14 seconds. Since time can't be negative, the tee shirt was in the air for approximately 4.60 seconds before hitting the ground.

Therefore, the answers are:

a. The tee shirt is approximately 32 feet high 1 second after being thrown.

b. The tee shirt was in the air for 4.5 seconds before it was caught.

c. The tee shirt was in the air for approximately 4.60 seconds before hitting the ground.