Question

Medal and Fan, I need help with last question.
The function H(t) = -16t2 + 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 90 feet with an initial velocity of 50 feet per second. Create an equation to find the time taken by the projectile to fall on the ground. (2 points)
Part A:
H(t) = -16t^2 + vt + s
H(t
Part B: What is the maximum height that the projectile will reach? Show your work. (2 points)

Answer 1

I already did part A. I need help with Part B.

Answer 2

@Conqueror

Answer 3

the maximum heightreached will be when the velocity is zero

Answer 4

I'm not sure I understand why? Because gravity will stop it at one point, which is the maximum point?

Answer 5

the velocity is zero at the moment particle reaches maximum height

Answer 6

Oh thank you, can you help me with the other parts?

Answer 7

Part C: Another object moves in the air along the path of g(t) = 28 + 48.8t 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] (4 points)

Answer 8

we haven't done part B yet
I'm not sure what we put for t

Answer 9

H(t) = -16t^2 + 90

Answer 10

I think I finished part B.

Answer 11

-b / 2a
-50 / 2(-16) = -50 / -32 = 1.56
H(t) = -16(1.56)^2 + 50(1.56) + 90
H(t) = 129.06 (approximately.)

Answer 12

because the projectile starts at 90 feet above the ground and lands 90 feet below that level so the time to maximum height is not the total time

Answer 13

oh - parabolic path
sorry I cant help you with this