after 1 second the rock is 243 feet in the air after 2 seconds, it is 452 feet in the air find the height in feet of the rock after 5 seconds in the air.
rock? Did you mean rocket?
Assuming it started from a height of k ft, with an initial velocity of v ft/se
height = -16t^2 + vt + k
when t = 1 sec, height = 243 ft --> 243 = -16 + v + k
v + k = 259
when t = 2, height = 452 ---> 452 = -64 + 2v + k
2v + k = 516
subtract:
v = 257 , subbing back into v+k= 259, k = 2
height = -16t^2 + 257t + 2
let t = 5, now find height = ...
To find the height of the rock after 5 seconds, we can use the information given and apply a pattern to calculate the height at each second.
Given:
- After 1 second: 243 feet
- After 2 seconds: 452 feet
To determine the change in height between the first and second second, we subtract the initial height from the final height:
452 feet - 243 feet = 209 feet
We notice that the change in height between each successive second is constant at 209 feet.
To find the height after 5 seconds, we need to calculate the change in height between the second and fifth second. Since the change in height per second is constant, we can multiply the change in height per second by the number of seconds:
Change in height = 209 feet/second × (5 seconds - 2 seconds)
Change in height = 209 feet/second × 3 seconds
Change in height = 627 feet
Finally, to find the height after 5 seconds, we add the change in height to the height after 2 seconds:
Height after 5 seconds = Height after 2 seconds + Change in height
Height after 5 seconds = 452 feet + 627 feet
Height after 5 seconds = 1079 feet
Therefore, the height of the rock after 5 seconds in the air is 1079 feet.