Projectile problems are presented along with detailed solutions.These problems may be better understood when projectile equations are first reviewed. An interactive html 5 applet may be used to better understand the projectile equations. Problems with Detailed Solutions. Projectile motion is the motion experienced by an object in the air only under the influence of gravity. A projectile, that is launched into the air near the surface of the Earth’s and moves along a curved path, or in other words a parabolic path, under the action of gravity, assuming the air resistance is negligible.

For our purposes, a 'projectile'is any object that is thrown, shot, or dropped. Usually the object ismoving straight up or straight down. An object is launchedat 19.6meters per second (m/s) from a 58.8 -metertall platform. The equation for the object's height sat time tseconds after launch is s ( t )= –4.9 t 2 + 19.6 t+ 58.8, wheres isin meters. When does the object strike the ground?What is the height (aboveground level) when the object smacks into the ground? Well, zero, obviously.So I'm looking for the time when the height is s= 0.

I'll set sequal to zero, and solve:0 = –4.9 t 2+ 19.6 t + 58.80 =t 2 – 4 t – 120 =( t – 6)( t + 2)Then t= 6 or t= –2. The second solutionis from two seconds before launch, which doesn't make sense in thiscontext.

(It makes sense on the graph, because the line crosses thex -axisat –2,but negative time won't work in this word problem.) So ' t= –2 ' is an extraneoussolution, and I'll ignore it.The object strikesthe ground six seconds after launch.Note the construction ofthe height equation in the problem above. The initial launch height was58.8meters, and the constant term was ' 58.8 '.The initial velocity (launch speed) was 19.6m/s, and the coefficient on the linear term was ' 19.6 '.This is always true for these up/down projectile motion problems. (Ifyou have an exercise with sideways motion, the equation will have a differentform, but they'll always giveyou that equation.)The initial velocity is the coefficient for the middle term, and the initialheight is the constant term.

And theon the leading term comes from the force of gravity. This coefficientis negative, since gravity pulls downward, and the value will either be' 4.9 '(if your units are 'meters') or ' 16 '(if your units are 'feet'). In general, the format is:s ( t )= – gt 2+ v 0 t+ h 0.where ' g 'here is the ' 4.9 'or the ' 16 'derived from the value of the force of gravity (technically, it's halfof the force of gravity, but you probably don't need to know that rightnow), ' v 0 '('vee-naught', or 'vee-sub-zero') is the initial velocity,and ' h 0 '('aitch-naught', or 'aitch-sub-zero') is the initialheight.Memorize this equation(or at least its meaning), because you may need to know this on the test. An object in launcheddirectly upward at 64feet per second (ft/s) from a platform 80feet high. What will be the object's maximum height? When will it attainthis height?Hmm.

They didn't giveme the equation this time. But that's okay, because I can create theequation from the information that they didgive me. The initial height is 80feet above ground and the initial speed is 64ft/s. Since my units are 'feet', then the number for gravitywill be 16,and my equation is:s( t)= –16 t 2 + 64 t + 80They want me to find themaximum height. For a negative quadratic like this, the maximum willbe at the vertex of the upside-down parabola. So they really want meto find the vertex. From,I know how to find the vertex; in this case, the vertex is at (2,144):h= – b / 2 a= –(64)/2(–16) = –64/–32 = 2k= s (2)= –16(2) 2+ 64(2) + 80 = –16(4) + 128 + 80 = 208 – 64 = 144But what does this vertextell me?

According to my equation, I'm plugging in time values and extractingheight values, so the input ' 2 'must be the time and the output ' 144 'must be the height. Copyright© Elizabeth Stapel 2004-2011 All Rights ReservedIt takes two secondsto reach the maximum height of 144feet. An object is launchedfrom ground level directly upward at 39.2m/s. For how long is the object at or above a height of 34.3meters?My units this time are'meters', so the gravity number will be ' 4.9 '.Since the object started at ground level, the initial height was 0.Then my equation is:s( t)= –4.9 t 2 + 39.2 t. AdvertisementSince this is a negativequadratic, the graph is an upside-down parabola.

I can find the twotimes when the object is exactly 34.3meters high, and I know that the object will be above 34.3meters the whole time in between. Why 'two time', and howdo I know that the time period is between those two times? Because thefirst time will be when the object passes a height of 34.3meters on its way up to its maximum height, and the second time whenbe when it passes 34.3meters as it is falling back down to the ground. So I have to solvethe following:–4.9 t 2+ 39.2 t= 34.3t 2– 8 t + 7 = 0( t– 7)( t – 1) = 0Then the object is at34.3 metersat one second after launch (going up) and againt at seven seconds afterlaunch (coming back down). Subtracting to find the difference, I findthat:The object is ator above 34.3 metersfor six seconds.Don't be surprised if manyof your exercises work out as 'neatly' as the above exampleshave.

Many textbooks still engineer their exercises carefully, so thatyou can solve by factoring (that is, by quickly doing the algebra). However,heavy dependence on calculators is leading more texts to create 'interesting'(that is, needlessly complicated) exercises, so some (or all) of yourexercises may involve much more messy computations than have been displayedhere. If so, study these 'neat' examples carefully, until youare quite sure you follow the reasoning. After the semesteris over, you discover that the math department has changed textbooks(again) so the bookstore won't buy back your nearly-new book. You andyour friend Herman decide to get creative.

You go to the roofof a twelve-story building and look over the edge to the reflectingpool 160 feetbelow. You drop your book over the edge at the same instant that Herman chuckshis book straight down at 48 feetper second. By how many seconds does his book beat yours into the water?Our initial launch heightswill be the same: we're both launching from 160 feetabove ground. And the gravity number, since we're working in feet, willbe 16.My initial velocity is zero, since I just dropped my book, but my buddyHerman's velocity is a negative 48,the negative coming from the fact that he chucked his book downrather than up. So our 'height' equations are:mine: s( t)= –16 t 2 + 160his: s( t)= –16 t 2 – 48 t + 160In each case, I need tofind the time for the books to reach a height of zero ('zero'being 'ground level'), so:mine: 0= –16 t 2 + 160, t 2 – 10 = 0,so t= ± sqrt(10)his: 0= –16 t 2 – 48 t + 160, t 2+ 3 t – 10 = 0, ( t + 5)( t – 2) = 0,so t= –5 or t= 2I will ignore the negativetime values. His book hits the water after two seconds, and mine hitsafter sqrt(10) seconds,or after about 3.16seconds.

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