Question

Help?

Answer 1

The function H(t) = -16t^2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second.

Part A: The projectile was launched from a height of 100 feet with an initial velocity of 60 feet per second. Create an equation to find the time taken by the projectile to fall on the ground.

Part B: What is the maximum height that the projectile will reach? Show your work.

Part C: Another object moves in the air along the path of g(t) = 20 + 38.7t where g(t) is the height, in feet, of the object from the ground at time t seconds.

Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values]

Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know?

Answer 2

Please anyone I'll fan and medal if that matters..

Answer 3

What lesson?

Answer 4

Not on a lesson, from a pack my teacher gave me to practice on

Answer 5

H(t) = -16t2 + vt + s
H(t) = -16t2 + (60)t + 82
since H(t) is the height and the time taken to hit the ground is the highest height x2
2(H(t)) = 2(-16t2 + (60)t + 82)
2(H(t)) = -32t^2 + 120t + 164

Answer 6

So then what? For part a?

Answer 7

This is how to find the velocity, and this is for part B

Answer 8

The maximum was my final answer

Answer 9

Well what about finding the velocity for part a?

Answer 10

I don't understand what your answer was..

Answer 11

Does this help? At all...

Answer 12

Sorry I was talking to my mom

Answer 13

Yeah man one second I'm just trying to process this through my brain lol

Answer 14

Processing...50% Complete...@LearningIsAwesome 's Brain couldn't complete command

Answer 15

100% Processed!

Answer 16

Process is complete! lol... Thanks man, can you delete that message explaining it I got it down and I don't wanna get in trouble from kind of having similar explanation as one on the interwebssss

Answer 17

Ok got it!

Answer 18

Thanks man! Can I tag you later if I come across something else?

Answer 19

Go ahead! But I may not be online so forgive me if so

Answer 20

Haha yeah I got you :)