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Click here to view ALL difficulties on take trip Word ProblemsQuestion 262848: the formula h=vt-16t^2 offers the height h in feet of an object after t sec. Here v is the early velocity the the object expressed in feet every second. Expect a toy rocket was introduced from the ground straight up with an early velocity that 64ft/sec. How countless seconds ~ launch was the rocket 48 ft over the ground.ok so below is what i have so far:using the formula and what is provided i have 48=64t-16t^2 therefore i have to make the a quadratic prefer this 64t-16t^2-48=0 then multiply by -1 and also put the in the correct order i get:16t^2-64t+48=016(t^2-4t+3)=016(t-3)(t-1)so i gain two positive answers. Once i placed t=1 into the formula favor this:48=64(1)-16(1^2) i get48=48but how can something introduced at 64ft per second only be 48 feet increase after 1 second? is my answer wrong or to be i lacking something in the physics of the problem?i hope my notation is clear. Any aid you can give would be appreciated. Discovered 2 solutions by dabanfield, stanbon:Answer through dabanfield(803) (Show Source): You can put this equipment on your website! What you have actually looks good. Also though the intitial velocity is 64ft/sec, after ~ one 2nd the rocket would be just 64 ft high in the lack of gravity. For this reason it makes sense the after one second, with gravity, it would certainly be much less than 64 feet high. After ~ two more seconds (that is at t=3) the rocket has actually reached that high allude and is descending. The rocket"s course is a parabola and also so there are two times as soon as the rocket is in ~ 48 feet, once on the way up and once top top the means down again. Price by stanbon(75887) (Show Source): You have the right to put this systems on her website! the formula h=vt-16t^2 gives the height h in feet of things after t sec. Right here v is the initial velocity that the object expressed in feet every second. Mean a toy rocket was released from the ground directly up through an early velocity of 64ft/sec. How plenty of seconds after ~ launch was the rocket 48 ft above the ground. Ok so right here is what i have actually so far:using the formula and also what is offered i have 48=64t-16t^2 therefore i need to make that a quadratic choose this 64t-16t^2-48=0 then multiply through -1 and put the in the correct order i get:16t^2-64t+48=016(t^2-4t+3)=016(t-3)(t-1)so i gain two positive answers. When i put t=1 right into the formula choose this:48=64(1)-16(1^2) ns get48=48but how can something introduced at 64ft per second only be 48 feet up after 1 second? is my answer wrong or am i absent something in the physics the the problem?i expect my notation is clear. Any aid you can provide would be appreciated. -------------------------the formula h=vt-16t^2 offers the elevation h in feet of things after t sec. Here v is the early velocity that the thing expressed in feet per second. Expect a toy rocket was introduced from the ground right up v an early velocity the 64ft/sec. How countless seconds after launch was the rocket 48 ft over the ground. Ok so below is what i have actually so far:using the formula and also what is given i have actually 48=64t-16t^2 so i need to make that a quadratic choose this 64t-16t^2-48=0 then multiply by -1 and put that in the correct order ns get:16t^2-64t+48=016(t^2-4t+3)=016(t-3)(t-1)so i obtain two optimistic answers. As soon as i placed t=1 right into the formula like this:48=64(1)-16(1^2) ns get48=48but how can something launched at 64ft per 2nd only it is in 48 feet up after 1 second? is my answer wrong or to be i missing something in the physics the the problem?---------Ans: heaviness is pulling the object ago to planet at the price of 16 ft.per second------------------------------------------------------------------------I expect my notation is clear. Any assist you can provide would be appreciated.

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The formula h(t) = vt-16t^2 offers the height h in feet of an item after t sec.How many seconds ~ launch to be the rocket 48 ft above the ground.----16t^2 + 64t = 48-16t^2 + 64t - 48 = 0-16(t^2 - 4t + 3) = 0t^2 - 4t + 3 = 0(t-3)(t-1) = 0t = 1 second ; t = 3 seconds-------------------------------------==================================================Cheers,Stan H. 