Asked by ju
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.
Answers
Answered by
Reiny
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 = ...
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 = ...
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