Answer:
Step-by-step explanation:
This is a physics parabolic motion question. It was the hardest chapter for my students to grasp onto because it involved motion in 2 dimensions. We have y dimension stuff and x dimension stuff to remember.
y-dimension
a = -9.8 m/sec/sec
Δx = -28
v₀=15 m/sec
Let's explain some of this before we move to the horizontal dimension, shall we?
Acceleration in the metric system is always going to be -9.8 m/sec/sec. Negative because gravity is always pulling down.
Δx is the displacement in meters. This value is also negative because the ball is going to end up below the point from which it is dropped.
v₀ is the initial velocity. We have to find the velocity in the y dimension, because it will be different from the velocity in the x dimension. This is where the angle comes in. The formula for velocity in the y dimension is
[tex]v_{0y}=v_{0}sin\theta[/tex]
Our angle is either -20 degrees or 340 degrees. They're the same thing.
[tex]v_{0y}=15sin(-20)[/tex] and
[tex]v_{0y}=-5.1303[/tex]
Now we can go to our displacement equation and solve it for t. The displacement equation is
Δx = v₀t + 1/2at²
and we will use the info we gathered in this dimension:
[tex]28=-5.1303t+\frac{1}{2}(-9.8)t^2[/tex] (the 28 is supposed to be negative here and below!) and
[tex]28=-5.1303t-4.9t^2[/tex]
Set this up as quadratic equal to 0 so we can factor and solve for t. Set it equal to 0 because when the ball is on the ground, it's height is 0.
[tex]-4.9t^2-5.1303t+28=0[/tex] and factor however you like and get that
t = -2.9706 and t = 1.9236
Since time can NEVER be negative, t = 1.9236 seconds.
We can use that time now in our displacement equation in the x dimension. In the x dimension, acceleration is always 0. Our angle becomes important to us again because remember that the horizontal velocity is not the same as the vertical velocity. The formula to find the horizontal velocity is
[tex]v_{0x}=v_{0}cos\theta[/tex] which for us is
[tex]v_{0x}=15cos340[/tex] which gives us that
[tex]v_{0x}=14.095[/tex] meters per second.
Going back to the displacement equation from above and filling in our x dimension stuff:
Δx = (14.095)(1.923) + 1/2(0)(1.923)²
which is nice to work with because to the right of the plus sign all goes to 0. If you haven't ever been made aware of this fact, I'm going to enlighten you: the displacement equation in the horizontal dimension is EXACTLY the same equation as the motion equation with which you are extremely familiar: d = rt
d = 14.095(1.923) and
d = 27.104 meters
That probably wasn't quite brief, but these are involved problems and require explanation each step of the way. Good luck in physics!!
