If the initial velocity is 100 m/s at 42.8 ◦ above the horizontal and the initial height of the target is 95.5 m , how high above the ground does the collision take place? The acceleration due to gravity is 9.8 m/s 2 .

Since it only mentions the height we only have to concern ourselves with the y direction, up and down, so that's nice. Also, acceleration due to gravity is -9.8 m/s^2

First we want the formula for the position of the object, for that we need the initial speed. We have an initial speed but we need the speed in the y direction. To do that we break the speed we have into its x and y components. Imagine a right triangle where the hypotenuse is 100 and the angle between that and the horizontal leg is 42.8 degrees. Notice that the legs of the right triangle are vertical and horizontal, they represent the y and x components of the speed respectively. Anyway, we want the y component, so the vertical. Relative to the given angle that is the "opposite" side, when we use SOH CAH TOA. Let me know if this doesn't make sense and I can draw it out.

Now, using trig, we can find the y component with SOH. so sin(42.8)=y/100. Make sure your calculator is in degree mode of course. we get

y = 100*sin(42.8) = 67.94 so the vertical speed is 67.94 m/s.

With that now we can use the displacement formula s = vi*t + .5a*t^2 if you don't remember this formula you definitely should have some kind of paper or something with all your physics formula. Anyway, we can plug in all values except for t, which is our variable.

s = 95.5

vi=67.94

a= -9.8

95.5 = 67.94t + .5(-9.8)t^2

Then if you subtract 95.5 from both sides you have a traditional quadratic equation you can solve with the quadratic formula, or completing the square or however you solve them. If you need help with this step let me know, or if something else didn't make sense.