WEBVTT
00:00:00.910 --> 00:00:11.180
A particle projected with a velocity 𝑢𝐢 plus 𝑢𝐣 metres per second from a fixed point 𝑂 in a horizontal plane landed at a point in the same plane 360 metres away.
00:00:11.580 --> 00:00:15.620
Find the value of 𝑢 and the projectile’s path’s greatest height ℎ.
00:00:15.650 --> 00:00:18.610
Take 𝑔 equals 9.8 metres per square second.
00:00:18.860 --> 00:00:22.270
When dealing with projectile motion, it can be really useful to sketch a diagram.
00:00:23.100 --> 00:00:26.590
We know that the motion of the particle will look a little bit like this inverted parabola.
00:00:26.810 --> 00:00:32.660
We’re also told that the horizontal and vertical components of the initial velocity are both equal to 𝑢.
00:00:32.870 --> 00:00:35.300
So, we can drop in a right-angle triangle as shown.
00:00:35.480 --> 00:00:39.150
We also know that the particle lands 360 metres away.
00:00:39.220 --> 00:00:43.730
So, that means the final horizontal displacement of the particle is 360 metres.
00:00:43.810 --> 00:00:46.000
We also have acceleration due to gravity.
00:00:46.030 --> 00:00:47.310
That acts downwards.
00:00:47.400 --> 00:00:53.340
So, in order to answer these questions, we’re going to use the equations of constant acceleration, sometimes called the SUVAT equations.
00:00:53.460 --> 00:00:58.160
But we’re going to separate these into the horizontal and vertical motion of the particle.
00:00:58.370 --> 00:00:59.890
We’ll begin with the horizontal motion.
00:01:00.100 --> 00:01:03.570
Initial velocity in the horizontal direction, we’re told, is equal to 𝑢.
00:01:03.870 --> 00:01:07.000
We also know that horizontally there is no acceleration.
00:01:07.120 --> 00:01:07.860
So, that’s zero.
00:01:07.980 --> 00:01:12.750
And this means that its horizontal velocity, until it reaches the ground at least, will always be equal to 𝑢.
00:01:13.150 --> 00:01:18.110
We’re going to use the final horizontal displacement to help us calculate a few other bits of information.
00:01:18.280 --> 00:01:20.190
And we know that’s equal to 360.
00:01:20.230 --> 00:01:22.580
We don’t yet know anything about the time.
00:01:22.720 --> 00:01:24.310
We’ll now consider the vertical motion.
00:01:24.540 --> 00:01:27.370
The initial velocity vertically is 𝑢.
00:01:27.440 --> 00:01:31.690
We’re told that acceleration due to gravity, which we’ve labelled 𝑔, is 9.8.
00:01:31.720 --> 00:01:36.300
Now, that acts in the opposite direction that we defined the vertical velocity to be in.
00:01:36.430 --> 00:01:38.400
So, we’re going to say that’s negative 9.8.
00:01:38.780 --> 00:01:41.970
We don’t yet know anything about the vertical velocity at a given time.
00:01:42.140 --> 00:01:47.000
But we do know that when the particle hits the ground, its vertical displacement must be equal to zero.
00:01:47.240 --> 00:01:49.380
And we don’t yet know the time at which this occurs.
00:01:49.410 --> 00:01:53.030
We do, however, have enough information to help us find the value of 𝑢.
00:01:53.330 --> 00:01:55.030
Let’s list those SUVAT equations.
00:01:55.120 --> 00:02:06.370
We have 𝑣 equals 𝑢 plus 𝑎𝑡, 𝑠 equals 𝑢𝑡 plus a half 𝑎𝑡 squared, 𝑠 equals a half 𝑢 plus 𝑣 times 𝑡, 𝑣 squared equals 𝑢 squared plus two 𝑎𝑠, and 𝑠 equals 𝑣𝑡 minus a half 𝑎𝑡 squared.
00:02:06.600 --> 00:02:09.050
We’re going to begin by looking at the horizontal motion.
00:02:09.170 --> 00:02:13.340
And we’re going to use the third equation, 𝑠 equals a half 𝑢 plus 𝑣 times 𝑡.
00:02:13.370 --> 00:02:16.890
This will tell us the value of 𝑡, the time that the particle is in motion.
00:02:17.260 --> 00:02:24.320
We get 360 equals a half times 𝑢 plus 𝑢 times 𝑡, or simply 360 equals 𝑢 times 𝑡.
00:02:24.450 --> 00:02:28.740
By dividing through by 𝑢, we find that 𝑡 is equal to 360 over 𝑢.
00:02:29.080 --> 00:02:31.430
This remains consistent in the vertical direction.
00:02:31.460 --> 00:02:34.110
Remember, we’re looking at the point that the particle hits the ground again.
00:02:34.300 --> 00:02:36.230
This time we’re going to use the second equation.
00:02:36.230 --> 00:02:38.680
We’re not really interested in the vertical velocity.
00:02:38.870 --> 00:02:43.190
We get zero equals 𝑢𝑡 plus a half times negative 9.8 𝑡 squared.
00:02:43.410 --> 00:02:46.220
A half times negative 9.8 is negative 4.9.
00:02:46.220 --> 00:02:49.070
And then, we can replace 𝑡 with 360 over 𝑢.
00:02:49.390 --> 00:02:50.860
These two 𝑢s cancel.
00:02:51.010 --> 00:02:59.170
We then add 4.9 times 360 over 𝑢 squared to both sides then multiply through by 𝑢 squared and divide through by 360.
00:02:59.250 --> 00:03:07.550
So, we get 𝑢 squared is equal to 4.9 times 360, which means 𝑢 is equal to the square root of 4.9 times 360, which is equal to 42.
00:03:07.800 --> 00:03:13.510
Now, here, we were only interested in the positive square root since we define the initial horizontal velocity to be positive.
00:03:13.740 --> 00:03:16.810
And so, 𝑢 must be equal to 42 metres per second.
00:03:17.040 --> 00:03:21.430
The second part of this question asks us to find the projectile’s path’s greatest height.
00:03:21.770 --> 00:03:26.690
Now, there are a couple of ways that we can do this, but we can redefine a few elements in the vertical direction.
00:03:26.870 --> 00:03:29.070
We’re no longer interested in the horizontal motion.
00:03:29.160 --> 00:03:30.940
We know that 𝑢 is equal to 42.
00:03:31.150 --> 00:03:33.130
The height we’re interested in is ℎ.
00:03:33.180 --> 00:03:39.060
We don’t know the time at which this occurs, although we could work it out by finding the total time taken and then halving it.
00:03:39.170 --> 00:03:46.470
Alternatively, we recognise that at the moment that the particle reaches its highest height, its velocity in the vertical direction is zero.
00:03:46.690 --> 00:03:49.380
And so, we find the equation which does not include 𝑡.
00:03:49.530 --> 00:03:51.920
It’s 𝑣 squared equals 𝑢 squared plus two 𝑎𝑠.
00:03:51.990 --> 00:03:58.450
Substituting what we know about our particle, we get zero squared equals 42 squared plus two times negative 9.8 times ℎ.
00:03:58.590 --> 00:04:01.260
42 squared is 1764.
00:04:01.260 --> 00:04:03.590
And two times 9.8 is 19.6.
00:04:03.650 --> 00:04:05.980
So, we add 19.6ℎ to both sides.
00:04:06.170 --> 00:04:09.380
And all that’s left to do is divide through by 19.6.
00:04:09.440 --> 00:04:11.030
That gives us a value of 90.
00:04:11.390 --> 00:04:13.880
And so, we can say that 𝑢 is 42 metres per second.
00:04:14.020 --> 00:04:17.760
And the projectile’s path’s greatest height ℎ is 90, or 90 metres.