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This was a question I found about projectile motion, the question was what's the bike's speed when it took off.

Using $S=ut + 0.5at^2$, the time taken to reach the ground is $0.505 s$, they used this time to find the bike's initial speed at launch (10m/0.505s) = 20m/s.

What I don't understand is how can we use the time(t) obtained from the vertical motion to find it's horizontal velocity? Aren't they 2 seperate things. I'm very new to projectile motion, so a simple concise analogy might be extremely helpful.

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The key thing to remember is that the velocity and the acceleration are both vectors so they have $x$ and $y$ components. That is we need to write the velocity as:

$$ \mathbf v = (v_x, v_y) $$

and the acceleration as:

$$ \mathbf a = (a_x, a_y) $$

And the position is the vector $(x,y)$. Now when we write the equation:

$$ s = ut + \tfrac{1}{2} at^2 $$

This is really a vector equation because $\mathbf s$, $\mathbf u$ and $\mathbf a$ are vectors. It's really two equations, one for the $x$ component and one for the $y$ component:

$$ x = u_x t + \tfrac{1}{2} a_x t^2 $$

$$ y = u_y t + \tfrac{1}{2} a_y t^2 $$

We know the motorcycle is moving horizontally to start with, so $u_y=0$. We also know that gravity acts vertically downwards, so $a_x=0$. This means our two equations simplify to:

$$ x = u_x t $$

$$ y = \tfrac{1}{2} a_y t^2 $$

This is why the problem splits neatly into two parts. The second equation tells us that the vertical distance $y$ depends only on the time and $a_y$, and since we know both $y$ and $a_y$ we can calculate the time. Then the first equation tells us that the horizontal distance $x$ depends only on $u_x$ and $t$. Since we know $u_x$ and $t$ we can calculate $u_x$, which is what the question is asking.

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You're 100 % right, but what should not be confused here is the notion of time. Time passes relative to nothing, except at relativistic speeds. The time it took the bike to clear 10 m from his initial position is the same as the time it took the bike to fall 1.25 m (0.505 s). As you can obviously see the trajectory, the point at which the bike touches the ground is horizontally 10 m away from the origin and 1.25 m below it, so if by falling 1.25 m the bike covers 10 m forward, then the bike was falling and moving horizontally at the same time. Consider you start eating pizza and playing videogame at the same time, then you stop doing both also simultaneously, is the time it took you to eat pizza different from the time it took to play videogame? I believe it's a colossal NO, it's the same.

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