Yo Yo Yo Whatup my Physicists,
I got a request to solve a particular Kinematics Problem so I thought I’d post it here for the whole wide world to see. Or like 6 people who actually click on this.
It’s a pretty nice problem involving some juicy kinematics; not too long at all and good to revise your fundamentals. Give the problem a go before looking at the solution:
Alright so hopefully you’ve attempted it. The right answers are 6.9 m/s and 2.4 m respectively. Hopefully you got that. If you didn’t, well I’ll show you my solution and framework of thinking when doing such a problem.
So, I think that the first thing you have to realize when you look at this problem is that it will definitely involve Projectile Motion. How could I tell? Well, for one, within the figure itself this is shown:
That’s a pretty big hint that you’re going to have to use that 60° angle to calculate some horizontal and vertical components.
So, let’s get started with solving (a):
They’re asking us for the maximum height reached by the stone. At max height, the velocity is always zero. So, we can say that our final velocity v, in this case, is going to be equal to zero. Hmm, let’s keep thinking. What other variables do we know? Well, we know that we’re throwing a stone upwards, against gravity. So our acceleration is just going to be the opposite of when a stone is thrown downwards. Hence it can be said that acceleration, a, = -10 m/s^2.
So we know a, we know final velocity v, and we’re solving for s (max height). This Suvat equation comes to mind:
All we need now is u, or the initial velocity. My first instinct when looking at this problem was to simply use u to be 8, the velocity that they gave us. But recall that this is a 3-point question and will likely not be so simple. Further, we know in the back of our mind there is a good chance we’re going to need to use projectile motion.
After thinking about it for maybe a minute, and playing around with the triangle that they provide us, it makes sense to think that we’re going to only need to use the vertical component of the velocity provided as u. I mean we are only throwing the stone upwards: the vertical component of the velocity is what is required. Here’s what I play around with now:
Now, we have all our variables and are ready to solve for s:
Woohoo. Part (A) done. Let’s do (B), which is I think easier.
The horizontal distance travelled by the stone that you are asked to find is in other words called the range. The range is really easy to calculate; you just to need to know the following formula:
Now, how do we find V horizontal? Well, we just use the exact same method we use to compute V Vertical. Here’s what you do:
Well, there we are. That wasn’t too bad I think. If you have any questions, feel free to comment and I’ll answer them ASAP. Also, email us at email@example.com if you ever have any questions related to the IB.