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A bomber releases a bomb while flying vertically upwards at a velocity of 1500 ft/sec at a height of 10,000 feet. a) How long after it is released will it take the bomb to reach the ground. b) Immediately after releasing the bomb the bomber flies away horizontally at the rate of 1200 feet/sec. How far away from the point at which the bomb strikes the ground is the plane when the bomb strikes?

For part B they multiply 1200 by 10 seconds to find how far its gone. However my question is if I see $1200 feet/sec$ as a velocity why can't I integrate it to get $s(t)$ an just put plug in 10 seconds in the equation for $s(t)$

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    $\begingroup$ $$\int_0^{10}1200\ dt = 1200\cdot (10-0) = 1200\cdot 10$$ So there's no difference. $\endgroup$ – user137731 Jul 20 '16 at 4:14
  • $\begingroup$ Where does 10 seconds come from? It takes a lot longer than that for the bomb to reach the ground. Did you mean 100 seconds? $\endgroup$ – Ross Millikan Jul 20 '16 at 4:36
  • $\begingroup$ Hey Bye_world. Why isn't it v(t)= 1200t and not just v(t)= 1200? $\endgroup$ – samuel Jul 20 '16 at 4:55
  • $\begingroup$ Ross Milikan no I meant 10 seconds. I found the answer in part A. $\endgroup$ – samuel Jul 20 '16 at 4:55
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    $\begingroup$ If we take $g=32f/s^2$, then the time the bomb takes to hit the ground is almost exactly $100$ sec. I bet the $10$ is supposed to be $100$. $\endgroup$ – robjohn Jul 20 '16 at 13:11
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For part A, the height above the ground is $h=-16t^2+1500t+10000$. Setting this to zero and solving gives $t=100$ seconds. For part B, the plane is traveling horizontally at constant speed. The distance from the starting point is then $1200t$ feet, so at $t=100$ it is $120,000$ feet away from the release point. You then need to compute the hypotenuse of the right triangle for the final answer

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