Mixed Problems: Set 2 (Vectors and Projectiles)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Two balls are thrown with the same speed: one at fifteen degrees above the horizontal, the other at seventy five degrees. And both land at exactly the same spot. So tell me: do they stay in the air for the same time? Hmm, same speed, same landing spot. Today we will settle it, with one small ratio. And look, this is Set two of our mixed problems: vectors and projectiles, mixed together, exactly the way JEE Main asks them. So, let's begin.

0:37Vector toolkit: the angle hides in the length

So, first the vector toolkit. The little arrow on top of a letter means a vector, so this is A vector, and the two straight bars around it mean its magnitude, its length. Plain capital A, without the arrow, is that same length, and this Greek letter, theta, is the angle between the two vectors. So, mod of A vector plus B vector equals the square root of, A squared plus B squared, plus two A B cos theta. The tick-shaped sign is the square root, and the small two up top means squared, okay? Now, subtraction. Flip B vector around, and add it to A vector. So tell me, for A vector minus B vector, what changes in this formula? Just one sign, that's right! Plus two A B cos theta becomes minus two A B cos theta. So mod of A vector minus B vector equals the square root of, A squared plus B squared, minus two A B cos theta. Clear? Now, how does JEE use this? Very often, the lengths are given and the angle is asked. Then square the formula to remove the root, and solve for cos theta. That's it, that is the whole trick.

1:44Projectile toolkit: T, H and R

Alright, now the projectile toolkit. Say a ball is thrown with speed u, at an angle theta above the horizontal, and g is the acceleration due to gravity. One condition: it is thrown from level ground and lands back at the same level. These formulas are for that case only, got it? Capital T, the time of flight, equals two u sin theta by g. The line means divided by, okay? Capital H, the maximum height, equals u squared sin squared theta by two g. Sin squared theta means sin theta, whole squared. And capital R, the range, equals u squared sin two theta by g. Three formulas, one angle. With me? Now tell me. Same speed, same g. Which other angle gives exactly the same range as theta? Ninety degrees minus theta, yes! The little circle means degrees. Theta and ninety degrees minus theta give the same sin two theta, so the same range. Keep this pair in mind. It is our puzzle from the start, right?

2:47Topper's trick: same speed? Take a ratio

Now a trick that toppers use in this set. Many questions compare two throws with the same speed u, and of course the same g. Only the angle changes. So, in the formula for the time of flight, what decides capital T now? Only sin theta! Exactly. Same u, same g, so T is proportional to sin theta. This sign, like a small alpha left open, means proportional to. The same way, in the height formula, only sin squared theta can change. So H is proportional to sin squared theta. Really simple. So in a ratio, u and g cancel completely. T one by T two equals sin theta one by sin theta two, and H one by H two equals sin squared theta one by sin squared theta two. The small one and two below mark the two throws, okay? You don't even need the speed! Got it?

3:42Example 1: angle between A and B if |A - B| = |A| = |B|

Now, Example one, an easy vector question from this set. Mod of A vector minus B vector equals mod A vector, which equals mod B vector. So all three lengths are equal. Find the angle between A vector and B vector. Pause the video, and try it first. Done? Okay, let's check. Many students grab the first formula from habit, the one with plus two A B cos theta. They get cos theta equals minus half, so one hundred and twenty degrees, right? No, no, careful! That plus sign is for the sum, not the difference. Here we have A vector minus B vector, so the cos theta term is minus. All three lengths are equal, so call each one capital A, okay? Then mod of A vector minus B vector, whole squared, equals A squared plus A squared, minus two A squared cos theta. But the left side is also just A squared, because the length of the difference is also capital A, right? So two A squared, minus two A squared cos theta, equals A squared. Divide everything by A squared. Two minus two cos theta equals one. So cos theta equals half. The double arrow means, therefore, and theta is sixty degrees! Look, one wrong sign would have given a completely different answer. Simple, isn't it?

5:03Example 2: time of flight and height ratio at 15° and 75°

Now a harder one, and it is our puzzle from the start! Two projectiles with the same speed are fired at fifteen degrees and seventy five degrees. Find the ratio of their times of flight, and of their maximum heights. Pause and try. Hint: use the topper's trick. Done? Okay, let's check. First, look at the two angles. Fifteen plus seventy five is ninety. So they are exactly the pair theta and ninety degrees minus theta, with the same range. Both land at the same spot, clear? Same speed, same landing spot. So both stay in the air for the same time, and the ratio is just one, right? No, no, careful! This is the JEE trap. Same range does not mean same time. By our trick, T goes with sin theta alone. So T one by T two is sin fifteen degrees by sin seventy five degrees. Not one at all! Now, seventy five is ninety minus fifteen, right? So what is sin seventy five degrees, in terms of fifteen degrees? Cos fifteen, yes! Exactly. So T one by T two becomes sin fifteen by cos fifteen, that is tan fifteen degrees. And tan fifteen degrees is two minus root three. That is the first answer. Now the heights. H goes with sin squared theta, so this ratio is tan squared fifteen degrees, the square of the first answer. Two minus root three, whole squared: four, minus four root three, plus three. So H one by H two is seven minus four root three. And there it is, puzzle solved! Same landing spot, but the low ball spends much less time in the air, and rises much less. Two lines of ratio did it all.

6:45Quick revision: vector lengths, ratios and same-range trap

So, quick revision. Vectors: lengths given and angle asked? Square the magnitude formula, and watch the sign of two A B cos theta. Projectiles with the same speed: T goes with sin theta, and H goes with sin squared theta, so take a ratio. And same range does not mean the same time of flight. Now go and try the practice problems on this page, then the whole set, with your clock running. You can do this, I promise. All the best!