Alright, I've got a Calc question because the answer to this problem isn't in the book and I'm positive a problem like this will be on the test.
A ball is launched upward at a speed of 160ft/sec from a cliff 176 feet above the ground.
a) find the velocity function v(t) and the position function s(t)
b) at what time does the ball reach its highest point? What is its maximum height above the ground?
c) when does the ball strike the ground? What is its velocity at that instant?
d) Set up the 2 definite integrals that would give the total distance traveled by the ball.
a) Well we know from the information given that V sub-zero is 160ft/sec and S sub-zero is 176ft. V'(t) which is equal to acceleration a(t) = -32feet/sec. So v(t)=-32t + 160
from that s(t) = -16t^2 + 160t +176
The 160 and 176 came from V sub-zero and S sub-zero and just solving for the constant C.
b) Ball would reach it's highest point when the velocity equals 0...so
0= -32t + 160
t=5sec
now I'd find s(5) which is 576 feet.
c) Ball strikes the ground when the position = 0, so 0=-16t^2 + 160t +176
After some factoring I got t=11 and t=-1, time can't be negative so I disregard the 2nd. To find the velocity at this point I use v(11) which is -192 , negative because it accounts for direction.
d) For this one I would use a sign chart to determine where the function is negative. So that I can figure out my intervals on my integrals. From earlier work we see that the function reaches 0 velocity at t=5 so I would use 5 on the number line like so:
<---------------->
0 5 11
v + -
from 0 to 5 velocity is positive, from 5 to 11 velocity is negative
so to set up the integral, I divide it into two integrals the first definite integral from 0 to 5, the 2nd from 5 to 11.
the first function from 0 to 5 would use the function -32t + 160 and the 2nd would use 32t - 160, since we want the distance traveled and not the displacement. Both would be added together to get the final result.
Correct me where I'm wrong here, this is how I'd do the problem.
A ball is launched upward at a speed of 160ft/sec from a cliff 176 feet above the ground.
a) find the velocity function v(t) and the position function s(t)
b) at what time does the ball reach its highest point? What is its maximum height above the ground?
c) when does the ball strike the ground? What is its velocity at that instant?
d) Set up the 2 definite integrals that would give the total distance traveled by the ball.
a) Well we know from the information given that V sub-zero is 160ft/sec and S sub-zero is 176ft. V'(t) which is equal to acceleration a(t) = -32feet/sec. So v(t)=-32t + 160
from that s(t) = -16t^2 + 160t +176
The 160 and 176 came from V sub-zero and S sub-zero and just solving for the constant C.
b) Ball would reach it's highest point when the velocity equals 0...so
0= -32t + 160
t=5sec
now I'd find s(5) which is 576 feet.
c) Ball strikes the ground when the position = 0, so 0=-16t^2 + 160t +176
After some factoring I got t=11 and t=-1, time can't be negative so I disregard the 2nd. To find the velocity at this point I use v(11) which is -192 , negative because it accounts for direction.
d) For this one I would use a sign chart to determine where the function is negative. So that I can figure out my intervals on my integrals. From earlier work we see that the function reaches 0 velocity at t=5 so I would use 5 on the number line like so:
<---------------->
0 5 11
v + -
from 0 to 5 velocity is positive, from 5 to 11 velocity is negative
so to set up the integral, I divide it into two integrals the first definite integral from 0 to 5, the 2nd from 5 to 11.
the first function from 0 to 5 would use the function -32t + 160 and the 2nd would use 32t - 160, since we want the distance traveled and not the displacement. Both would be added together to get the final result.
Correct me where I'm wrong here, this is how I'd do the problem.
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