 Now we're going to look at another problem involving projectile motion. Your friend throws a dodgeball at you, but luckily you managed to dodge it. The dodgeball bounces off the ground with a speed of four meters per second at an angle of 40 degrees to the horizontal and hits a wall one meter away. How fast is the ball travelling when it hits the wall? There's a lot of information here, so the first thing to do is to draw a diagram to help keep all that information close by. The only part that we care about is what happens after the dodgeball bounces off the ground, so that's what we'll draw. We know that the wall is one meter away. We know the initial speed was four meters per second and it was travelling at an angle of 40 degrees to the horizontal. To solve this problem, it's easiest to work out the horizontal and vertical components separately when the dodgeball hits the wall. Then we can add the vector components back together and get the total velocity. We start by writing an expression for our vertical and horizontal velocities, which we can get from trigonometry. We find that vix is equal to 4 times cos 40 meters per second, while viy is equal to 4 times sin 40 meters per second. Now, the wall is one meter away from where the ball bounced. The horizontal velocity vx doesn't change, it's the same at the wall as it was initially. The vertical velocity of the wall is given by g times t plus viy, where t is the time it takes to reach the wall, since there's a constant acceleration provided by gravity. So, before we add the vertical and horizontal components together, we need to figure out what t is. This is fairly easy to work out from the horizontal motion. Since the horizontal velocity is constant, the horizontal distance x is equal to vix times t, which we can rearrange to find t. So now we can substitute this into our expression for the final vertical velocity. And substituting in our values, we find that the vertical velocity is equal to minus 0.62 meters per second. The minus here means that the ball is moving downwards. Doing the same for the horizontal velocity, we multiply vix by t and find that the horizontal velocity at the wall is 3.06 meters per second. So now we just need to add our vector components back together. Remember that because these are components of vectors, we can't just add the horizontal and vertical speeds. Instead, since they're at right angles to each other, we can add them together using Pythagoras's theorem. Thus, the total magnitude of the velocity when the ball hits the wall is the square root of the vertical velocity squared plus the horizontal velocity squared. And this comes out to 3.12 meters per second. So now we're almost there. Remember that velocity is a vector, so it needs a direction too. We can use trigonometry. The tan of an angle is the opposite over the adjacent, so the arc tan of the opposite over the adjacent will give us the angle. Putting in the different velocities, we get negative 11.5 degrees. The negative in the angle means that it's below the horizontal instead of above, which is what we expect based on our problem. So in summary, the ball hits a wall with a velocity of 3.12 meters per second, 11.5 degrees below the horizontal.