At a fourth of July celebration, a rocket is launched with an initial velocity of 125 feet per second. The height (h) of the rocket in feet above sea level is modeled by the formula h=125t-16t(squared), where t is the time in seconds after the rocket is launched.
a. What is the height of the rocket when it returns to the ground?
b. how many seconds is the rocket in the air before it returns to the ground?
c. About how high does the rocket actually reach?
2007-10-04 06:37:24 · 4 answers · asked by taylor_travers098 2 in Science & Mathematics ➔ Mathematics
a) 0
b) 125t - 16t^2 = 0
t(125 - 16t) = 0
so t=0 or 125 -16t = 0
125 = 16 t
125/16 = 7 13/16 secs
c) 125 -32t = 0
125 = 32t
125/32 = 3 19/32 secs= t
2007-10-04 06:49:32 · answer #1 · answered by timemccormick 3 · 0⤊ 0⤋
a) When the rocket returns to the ground, its height is clearly zero feet (hence the fact that it has returned to the ground).
b) Here we must make an assumption about the rocket's interaction with the atmosphere. The easiest thing to do is to assume there is no drag (as your original equation already appears to do). We must also assume that the rocket accelerates to 125 feet per second in no time at all and thus leaves the ground at this rate; another point is that we must also assume the rocket motor burns for zero seconds after the launch & thus commences deceleration immediately.
The rocket will decelerate at 32 feet per second per second, and so it will reach its greatest height (and thus also zero speed) after 125/32 seconds, then it will take the same amount of time to return to the ground. The total flight time will thus be 250/32 seconds, or 7.8125 seconds. Half of this time (i.e. 3.90625 seconds) will be spent going up, the other half coming back down.
c) Since the even-paced deceleration of the rocket will result in its average upward speed being exactly half of its initial upward speed, the average is thus 125/2 = 62.5 feet per second. Using the upward portion of the flight time as calculated in b) above, the height achieved would be 3.90625x62.5 = 244.140625 feet. Since this is ludicrously accurate for a rocket which will have a certain size to it, perhaps the best estimation would be to the nearest whole foot - i.e. about 244 feet.
I hope this helps, but please feel free to drop me a line if you'd like to discuss this or any other figure-work further.
2007-10-04 14:35:30 · answer #2 · answered by general_ego 3 · 0⤊ 0⤋
a. The height of the rocket when it returns to the ground = 0.
b. 125/16 = 7.1825 seconds = time in the air
c. max height when t = 125/32 so h = 244.4 feet
2007-10-04 13:56:13 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
a. 0
b. 7.8125 sec
set 0=125t - 16t^2
factor 0=t (125-16t)
t=0 and t=125/16= 7.8125 sec
c. 244.1406 ft (at t=3.906 sec)
dh/dt = 125-32t
t=3.90625 (max height at t=3.90625 sec)
2007-10-04 13:53:34 · answer #4 · answered by skipper 7 · 0⤊ 0⤋