Motion under Gravity

Transcript

0:00Introduction

Hello everyone! Now, one small puzzle before we start. Here is a balloon, rising at five metres per second. At thirty metres above the ground, it lets go of a packet. Now tell me, does the packet start falling straight down? Of course, right? It is just let go! No, no, wait. That is exactly the trap. By the end of this video, you will know where that packet really goes first, and when it lands. And look, JEE often asks this topic: towers, balloons, lifts and dripping taps. So, let's begin.

0:37Thrown straight up with speed u

So, first, one rule for this whole topic: we take up as positive. In the figure, the dark arrow with the plus sign points up, okay? Gravity pulls down, so the acceleration, a, is minus g, where g is the acceleration due to gravity. And it stays minus g the whole time, even at the top. Clear? Now we throw a ball straight up with speed u, the orange arrow, from the top of a tower of height h. It goes up, slower and slower. So tell me, at the very top, what is its velocity? Zero, exactly! For one instant, it stops. So the first equation of motion gives zero equals u minus g t, and t equals u by g: the time of ascent. And how high does it go? The third equation of motion, with v zero and a minus g, gives zero equals u squared minus two g capital H, so capital H, the maximum height above the point of throw, equals u squared by two g. The figure writes it H max; the small max just means maximum, okay? Alright, going up takes u by g. Now think: how long does the way down take? The same, u by g! See, play the way up backwards, like a video in reverse: it is just a fall from rest from the top, right? So the time of flight, capital T, back to the starting point, is two u by g. And when the ball comes back past the top of the tower, how fast is it moving? Yes, speed u again, just pointing down; the figure says speed u down. In fact, it passes every level with the same speed going up and coming down. Completely symmetric, and it saves a lot of time.

2:27Dropped from a height h

Next, we simply drop a body from a height h. Dropped means let go from rest, so u is zero. The ground is h below the start, so the displacement is minus h, okay? The second equation of motion gives minus h equals minus half g t squared. The minus signs cancel, so t equals the square root of two h by g; that long tick is the square root sign. And the speed at the ground? The third equation gives v squared equals two, into minus g, into minus h. Minus into minus is plus, so v is root of two g h. Got it? Alright, now tell me. Does a falling body cover the same distance every second? No, no! It keeps speeding up, so every second it covers more. The distances in successive seconds are g by two, times one, three, five, and so on; the three dots mean the pattern goes on. The odd numbers, in order! Beautiful, isn't it? Topper's trick: with g as ten, g by two is five. So the first second covers five metres, the next fifteen, and the third twenty five; the m means metres. Keep this in your head, okay?

3:41Thrown from a height: one equation

Now a very common JEE setup: a ball thrown from the top of a tower of height h, the grey block in the figure. At the ground, the displacement is minus h. So minus h equals u t, minus half g t squared. Look, one equation for the whole motion, even the part below the starting point. No need to break it into pieces. And the sign of u? Choose it by the direction. Thrown up, u is positive; thrown down, u is negative. That's it! Now, here is a trap. Two balls, same speed u, one thrown up and one thrown down from the same tower. Which one hits the ground with more speed? The one thrown down, obviously, right? No, careful! Both hit the ground with the same speed. Look at the figure: the ball thrown up comes back to the top with speed u, going down. From there, it is just like the ball thrown down. So either way, the ground speed is root of u squared plus two g h. With me? One last rule. Something released from a moving balloon or lift starts with the velocity of the balloon or lift. Remember our packet? We will settle it in Example two.

4:52Example 1: stone dropped from 45 m, time and speed

Time for Example one, an easy one. Here, a stone is dropped from a height of forty five metres. Find the time of fall, the speed at the ground, and the distance covered in the last second. Take g as ten metres per second squared; that is the m s to the power minus two. Pause the video, and try it first. Done? Okay, let's check. Time of fall: root of two into forty five, by ten. That is root nine, three seconds; the s means seconds. Speed at the ground: root of two into ten into forty five, root of nine hundred. Thirty metres per second; the m s to the power minus one means metres per second. Now the last second. Forty five metres in three seconds, so fifteen metres in the last second, right? No, no, careful! It keeps speeding up. So find the distance in the first two seconds: half, into ten, into two squared. Twenty metres. The last second covers the rest: forty five minus twenty, twenty five metres. And check with our topper's trick: five plus fifteen plus twenty five is forty five, exactly the height! Perfect match.

6:12Example 2: packet released from a rising balloon

Now our puzzle from the start, JEE level. Here, a packet is released from a balloon rising at five metres per second, when the balloon is thirty metres above the ground. When does the packet reach the ground? Take g as ten again. Pause the video, and try it first. Alright, first the starting velocity. The packet is just let go, so u is zero, right? No, no, careful! This is the trap from our puzzle. The packet was moving up with the balloon, so it starts with the balloon's velocity: plus five metres per second, upwards. So it does not fall straight down at all. It first goes up a little, then comes down. Mystery solved! Now our one equation. The ground is thirty metres below the start, so minus thirty equals five t, minus five t squared; half g is five. Clear? Divide by five and bring everything to one side: t squared minus t minus six equals zero. Now you factorise it. Quickly! Yes, t minus three, into t plus two. So t is three seconds. And minus two? Negative time means before the release, so we reject it. See? The right starting velocity, one equation, and the question is done.

7:33Quick revision: max height, time of flight, free fall

So, quick revision. Up positive, a is minus g; thrown up, capital H is u squared by two g, and capital T is two u by g. Dropped: t is root of two h by g, v is root of two g h, and the distances in successive seconds go one is to three is to five. And from a balloon or lift, start with its velocity. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!