Question

2. An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. What will be the object's maximum height? When will it attain this height? When will it hit the platform again? If it passed the platform and hit the ground (0ft), how long will it take?

Answer 1

@horsegirl27 help me please

Answer 2

what do you know about trajectory equations?

Answer 3

if this is calculus based, we can start with acceleration and work our way up to velocity and position equations

Answer 4

Use equations of motion. acceleration will be g. initial speed will be 64. at turning point, final speed will be 0

Answer 5

this is alegbra two and how do i set up this forulma

Answer 6

@amistre64 I'm kinda busy, you hold the fort :D

Answer 7

algebra 2 eh, which deals with quadratics of memory serves.

when time =0 we are at 80 so we have some:

y = a t^2 + bt + 80

im not sure how to explain why the b value is our starting velocity without using calculus ... but that gives us:

y = a t^2 +64t + 80

we arent given any reference points to start with .... calculus would help us out immensely on this

Answer 8

as long as the site behaves i spose i can hold it :)

Answer 9

using calculus

the only force acting upon the object is gravity: so its acceleration is -32 feet per second squared

a(t) = -32

since acceleration is the rate of change in velocity, we can integrate it to get:

v(t) = -32t + C

given that v(0) = 64, C=64: v(t) = -32t + 64

now velocity is defined as the rate of change of position, so we can integrate again to get

h(t) = -32 t^2/2 + 64t + K

given that h(0) = 80, K=80: h(t) = -16 t^2 + 64t + 80