can i please get some help with this projectile motion
Sure! You might have to read this a few times - This projectile question is nuts, I remember doing it in my HSC year.
Let's not put numbers in until the end. So we know it reaches a maximum height of \(2h\) (double the pole) - I'm assuming you can derive the projectile motion formulae. This happens when the vertical velocity is zero:
We substitute that value into the formula for vertical displacement:
So that's one formula. NOW, let's consider when it gets to the height of the bird, \(y=h\), and use this relationship:
This is a quadratic equation in terms of \(t\), which we can solve with the quadratic formula. You'll get this:
}{g})
Let's take stock,
that is the time of flight before it reaches the height of the pole. The higher of these is when the bird is struck, because the bird is struck on the way down. Keep in mind also that the bird is travelling 10 metres per second to the right through the time of flight. We can obtain a distance expression:
}{g})
Now keep in mind that the rock was thrown
to hit the bird. So, it also reaches a height of h at the point where the bird started. Meaning, the time taken for the rock we threw to travel the distance travelled by the bird is just the difference between the two roots of the quadratic (since each corresponds to a time of flight to reach y=h):
We have a horizontal distance travelled and a horizontal time taken, we can use that to find horizontal velocity (since there is no horizontal acceleration):
}{g}\times\frac{g}{2\sqrt{gh}}=5\sqrt{2}\left(\sqrt{2}+1\right))
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