I don't get escape velocity.  Couldn't you just go 1 mph up until you were out?
604528
Gravity won't let you. You have to go into orbit first, and 1 mph will not do that. You have reach escape velocity to escape earth's gravity, which is 11.2 km/s.
943563
I have no clue of any, but I know that the escape velocity of Earth (the speed of which it takes to escape Earths gravity) is 25,000 mph. Try launching six and a half tons 25,000mph!=)
895548
Minimum speed for what?\nIf you want to leave earth orbit then you need to be over the earth escape velocity.
747360
Well gosh....  \n\nsince the only linear unit mentioned in the question is 18 ft/s  let's assume that the 60 is going to give us acceleration in ft/sec^2 \n\nSo, the position function for this is the second integral of the acceleration function, right? \n\nso...  the first anti-derivative of 60t is  30t^2   which gives the velocity, so at the end of the boost phase, our rocket is moving 30 * 3 * 3 = 270 ft/s \n\nand the second antiderivative is 10t^3  right?  \n\nSo, the position at time T is 10 * 3 * 3 * 3 = 270 feet.  (odd coincidence, but there you are...) \n\nNow, the rocket begins a SECOND acceleration by gravity of -32 ft/s negative 'cause we're accelerating DOWN now... ....   but this time the acceleration is constant,  so \n\nV = gt   and s = 1/2  a  t^2 so, we're slowing from 270 feet per second to zero at the top...  and  t = v/g = 270 / 32 = 8.44 seconds and the height we coast to is the 270 we had at the end of boost plus   1/2  *  32  * 8.44 * 8.44 = 1140 feet.  \n\nThe parachute is going to open after 14 seconds...   so, 14-8.44 is 5.56 seconds of free-fall (and we'll ignore air resistance again) so, we go from zero to something in 5.56 seconds....   and position is 1140 - ( 1/2  * g  * t^2 ) again...   so, the parachute opens at \n\n1140 - (1/2  * 32  * 5.56 * 5.56 )  = 645 feet. \n\nad the velocity is  gt = -32 * 5.56 = -178 feet per second...  \n\nNow, for the NEXT five seconds, we slow under the effect of the parachute   to  -18 ft/sec -- so,  we slow by 160 feet per second in five seconds, so the acceleration of the parachute is (oddly) 32 feet/second squared (huh, how odd is that? )  \n\nSo, the distance traveled during that phase is Vo t + 1/2 at^2  \n\nSo, we have (-160 * 5) + (1/2*32*5*5) =  -800 + 400 = 400 feet.  \n\nand from 400 feet up, we continue down at 18 ft/s for an additional  22 seconds...  \n\nand the rocket lands at t = 22 + 5 +  14 + 3  = 44 seconds \n\nAnd yes, I dropped a bunch of decimal points.  Your original data was only to two significant digits....
145934
You need to go from 0 to 11.2 km/s (earths escape velocity) and you need to overcome earths gravitational pull, which is 9.82 N/KG, in order to do so. You are constantly being accelerated downward with one G (9.92 m/s^2). If you would create a force upwards of one G you would hover (0 G). Let´s say you had a rocket engine that could thrust your spaceship upwards with 2 Gs. How long would it take to reach escape velocity? You use the equation t=v/a where v is 11200m/s and a is 9.81m/s^2 and get 1141.7s. You need to accelerate for 19 minutes to reach escape velocity. Would be sweet if you could do that but it is a long time to keep an engine running. Let´s accelerate twice as fast. Put a as 20m/s^2 and you´ll get 560 seconds or 3.3 minutes. \nOK. So now we´re cooking. How much thrust do you need to get this kind of acceleration? It´s another simple equation: F=ma. We have a (20m/s^2). But you also need m that is the mass of the spacecraft. This we can only guess but lets say a tonne (1000 kg). Then it is pushing downwards onto the launch pad with the force of 1 G wich is 9810 N (9.81x1000). We need to overcome this by adding another G to our thrust so a is 3Gs or 29.81m/s^2 (20 + 9.81). Thus we get 29810 N of thrust for 560 seconds to kick that 1 tonne payload into outer space. That´s 40.5 horsepowers to you non-metric folks (go metric already. A pound isn´t alwasy a pound is it? So measuring thrust in pounds isn´t valid). "But wait. Doesn´t that sound a little too sweet? 40.5 hps? My old Dodge can do better than that". That is correct. You also need fuel to make thrust. The fuel adds weight to the spacecraft which means we need more thrust. Whch means the need for more fuel. Also the craft is exposed to alot of drag from the air. More fuel. More weight. The farther you get from earth the less the value of a G gets but the spacecraft uses up fuel which makes up for the amount of force needed for acceleration. And so on. So this little thought experiment can only take us so far but I´ll stop here anyway. Now you got an idea atleast.
444555
No one knows for sure where either of these took place.  As for minutes.. I guess it would depend on how fast you are moving, what direction you travel, and by what method.  Why would you care about this anyway?  If this is a joke, what is the punchline?
412533
this is the concept of terminal velocity.  The cat will reach a certain speed and then go no faster because the friction caused by the air will counteract gravity to a certain extent.
133306
a. it is not the speed you "must" go because it is a maximum safe limit, you can easily go 50 if the limit is 55.\nb. gives the highest speed you "should" go.  not "can" go because you can easily o faster that the speed limit but you should't.\nc. it is not there to say "hey go less than this or get a ticket." it is there to show you what the safest speed would be.
557602
This is how you solve it. Use dimmensional canellation, but you will have to do it twice and subtract the difference\n\nA. (5 miles/1)  *   (1 hr/55 mi)  * (60 min/1 hr)\n   =5.45 minutes\n \n   \nB. (5 miles/1)*  (1 hr/65 miles) * (60 min/1 hr)\n   =4.6 minutes\n    \nYou saved 1.2 minutes. Not much at all. Which is what they say, speeding only saved 2 or so minutes in a 15-20 minute trip. BUt it tears the bejesus out of your engine
342841
It is different for every different circumstance. The key is when the force of gravity pulling you down is exactly equal to the drag on you from the rushing air, you arrive at terminal velocity. If you have a lot of drag (parachute deployed, a lot of loose clothing), it happens pretty quick. If you have low drag (tights, bald) then it takes longer.
743381
Velocity(at time t) = Velocity (at time = 0) + Acceleration x Time\n\nV(t) = Vi + At\nV(t) = 32 ft/sec + (-32 ft/sec/sec)(t)\nV(1 sec) = 32 - 32 = 0 ft/sec\n\ny(t) = y(initial) + (V x t) + 1/2(A)(t^2)\ny(1 sec) = 0 + (32) +  (-16) = 16 feet\n\nTo hit ground, y = 0, so (V x t) = 1/2(A)(t^2)\n32t = 16t^2\n2 = t, t = 2 seconds\n\ny(max) occurs at zero velocity, which is at t = 1 sec, x = 16 feet
21030
Assuming you have a tunnel that goes through the earth, take aside the fact you would survive the temperatures and such at the center, assuming there's no resistance (vacumn like environment), you would fall into the hole... accelerate to terminal velocity...pass the center, and slowly decelerating, until you reach back a speed of 0, where you will be at the other end of the hole, whereapart the whole process would start all over again unless you grasp the edge :)\n\nConsidering that the fact there's air, you'll encounter resistence, meaning you will accelerate to the center, start slowing down, but fall short of reaching the other end, then you will fall to the center again, fall even shorter of your original jumping position, and you'll oscillate back and forth until you indeed get stuck in the center of the earth.
730814
assuming that you could do it and you didnt die and all that, you would go through to the other side.  if you include air resistance, you wouldnt make it to the other side, you would accelarate until you reached the center of the earth or until you reached terminal velocity, then you would start slowing down after you passed the center.  you would then start falling back toward the center in the opposite direction, passing it again, you would oscillate back and forth like this, losing some distance from center on each pass, eventually you would come to rest at the center
399954
SPEED=dy/dt=distance/time.its not exactly something you invent.it was an axiom,a basis of understanding.
564019
Actually, velocity comes from having a strong core, hit the weights, work your abs, lower back, glutes, and thighs. A strong core will give you an increase of at least 3-5 mph on your fastball. Keep your arm loose too, and let your legs push you off the rubber. Trying to overthrow causes you to lose speed.
755432
About the same as the escape velocity for earth. (9.47^2/95.2)^0.5 = 0.97\n\nBest of Luck - Mike
391419
618 km/s (just under 1.4 million mph).  Important to know this figure if you want to move around in the solar system as it scales with your distance from the sun.\n\nYou can find the escape velocity of any object:\nEscape Velocity = SquareRoot((2GM)/r)\nG = Neton's gravitational constant\nM = mass\nr = radius