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Projectiles

Watch this to understand the theories of projectiles in Higher Physics (and score the perfect hoop-shot)

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The most beautiful game in the world? That’s easy: basketball.

The energy, action, aggression, all surging around the court; attackers and defenders trying to outdo each other, and then, from nowhere, the player with the ball shoots and the laws of physics take over.

The ball soars through a precise high arc towards the hoop in a perfect example of projectile motion.

With no obvious time for calculation, the player has instinctively understood the horizontal distance that needs to be covered - the NBA 3-pointline is roughly 23 feet from the basket - and the height gain and loss needed to go from the hand, upwards and then back downwards to land in a basket 10 feet off the ground. This calculation takes place in a fraction of a second while moving at speed and also trying to out-wit the defender.

If the player can get this calculation of the vertical and horizonal planes correct, then the ball will pass through the hoop and make the perfect whoosh sound and score points.

Easy, right? Hmmm....Let’s see how Physics makes sense of this.

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When we consider horizontal motion we assume there is no air resistance and after we release the ball there are no forces acting on the ball so it continues at a constant speed. Denoted by this rather boring graph and because the graph is so boring we can simply use the equation v=d/t to do calculations in the horizontal direction.

If we think about a thrown ball, vertically, we know that the object must always be accelerating downwards at 9.8 m/s/s. On a velocity time graph this will look like this. The upper-left portion of the graph represents the ball slowing down after it has left the hand.

When it reaches the top of its arc it must have a velocity of zero before it starts to accelerate back towards the ground. Whilst it looks like the ball is still moving in the same direction because the line has not changed it is in fact now moving downwards, we can tell this as the velocity is negative and therefore in a different direction.

So, to summarise the ball is thrown up with some initial vertical velocity (arrow pointing to graph) and gradually slows due to gravity until it reaches the top of its arc here (point at line meeting x-axis) before accelerating down again here.

Because the ball is accelerating our calculations here will require one of several possible equations. We call these the equations of motion or sometimes SUVAT because those letters denote our key quantities. When trying to figure out what is happening vertically the best thing to do is write SUVAT on your page and fill in the values you already know. Remember that a will be 9.8, if we only look at the ball on the way up v will be zero as it is about to change direction or u will be zero if we only think about the second half of the motion. And you will be given values for at least one of, s,t and u or v.

For example if I put some numbers on my graph (initial speed 19.6) and want to know how long it takes to reach its peak, I write u=(initial speed 19.6), v=0 as that is when it reaches its peak, and a=-9.8 as it always is. Then I see that I have 3 out of 4 for v=u+at sub in 0=19.6+(-9.8 x t) rearranging for t -19.6/9.8 = t gives me t = 2 seconds. Quick tip if you get a negative time your acceleration value should be negative.

If you can remember these key assumptions and apply them carefully to projectile questions you will quickly find yourself getting the right answers and scoring well. You might also get better at basketball, but no promises.

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