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Now, lets see one more application of the formulas developed in the previous article.Most of you must have come across the problemTwo projectiles have the same range R and heights H1 & H2 . Prove that R = 4 H1H2 Those of you who haven't solved this before try to do it. After you have broken your heads, or managed to solve it, come and see this solution.As the ranges are same, they are projected at say, and 90 - 4H1 = Rtan4H2 = Rtan(90-) => 4H2 = Rcotthen,16H1H2 = R2 R = 4 H1H2
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submitted by elessar_iitkgp (2220 m) on 19 May 2007 10:19:40 IST (2 comments 354 views)
Most of you have read projectiles and know how to find the equation of trajectory when the initial velocity, angle of projection are given. Lets consider a reverse problem: Given a general quadratic equation, lets find its initial velocity, angle of projection, time of flight, etc., etc, etc .... Consider a quadratic equation y = ax2+bx+c -------------(1) The X axis is considered to be the horizontal plane, and the Y axis is vetial, positive up. Now, for this equation to represent a projectile, it should be a parabola that opens downwards, ie, a<0 Also, it should have two real roots. For the case of a single real root, the equation represents a particle thr
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submitted by elessar_iitkgp (2220 m) on 18 May 2007 19:30:58 IST (2 comments 510 views)
We all know the standard formulas for range, height and time of flight of a projectile. Consider a problem like thisA particle projected obliquely at 45 deg has a range of 16m. Find the maximum height reached by the particle.Standard procedure:Write the formula for RangeEvaluate u.Find H.Time taken: 2 minLets do summa new.R = u2 sin2/ gH = u2 sin2 / 2gR/H = 4/tan4H = RtanUse this relation4H = 16 tan45H = 4Time taken :5 sYou get the idea. Similarly by doing R/T2 , H/T2 we get the following set of formulas4H = RtanRtan =(1/2)gT2 H = (1/8)gT2 Now try solving these problems:1. The range of a particle thrown at 45 deg is 8m. Find the time of flight.2. The maximum
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submitted by elessar_iitkgp (2220 m) on 19 May 2007 10:11:54 IST (0 comments 470 views)
