The problem tells us that the ball had an initial speed of 15.0 m/s when it was kicked at an angle 15° above the ground. Therefore, we can draw a line at a 15° angle above the horizon and write above it that the motion is at 15.0 m/s. This line should have an arrow at the right end of it to show that the motion is headed in that direction. From this, we can find the vertical and horizontal components of the motion later on. We can also make qualitative position (x) and velocity (v) graphs for the vertical and horizontal parts of the motion separately.
HORIZONTAL:
Horizontally, the ball moves to the right at a constant positive velocity, so therefore the line on the velocity (v) graph is horizontal and above the x-axis. To show this on the position (x) graph, we make a positive line that is straight but diagonal since the ball is moving to the right as time goes on with a constant speed.
VERTICAL:
Vertically, the ball lifts off the ground momentarily (since it was kicked at an angle of 15.0° above the horizontal), but then comes back to the ground when gravity pulls it down. Therefore, its position graph is in the shape of an upside down parabola, starting and ending at 0 m on the x-axis (0m represents the ground level). As far as velocity goes, since the ball lifts off the ground, it starts with a positive velocity, yet it ends up being negative as the ball is being drawn back to Earth. This means the line will start at 0 m/s on the v-axis and end up below the t-axis by the end of the motion.
Now that we have gotten the motion diagrams out of the way, we can do the necessary math to find the answers.
After plugging what we know into the formula, we are left with a Vy of 3.882 m/s. While we are at it, we should find the velocity in the x-direction as well. So, we will use the Pythagorean Theorem to find that since we know Vo and Vy.
We then find that the velocity in the x-direction is 14.489 m/s. Now that we know the angle and velocity in each direction, we can proceed to the next step: finding the time. We must solve for the time using the formula change in y equals 1/2 times gravity times time^2 plus initial velocity in the y-direction times time plus initial Y (Δy = 1/2*g*t^2 + Voy*t + Yo) since we have numbers for all the variables except for time. Once we do that, we are left with a quadratic equation, so we must use the quadratic formula to solve for t. For online purposes, I will signify the square root sign by raising the quantity to the 1/2 power.
(a) Now to find the maximum height, we will want to find the vertex of the parabola on the position in the y-direction graph. This is because the height is reached in the y-direction, and since the parabola is falling downward like the ball, the vertex would be the highest point that it attains. Using the formula negative b divided by 2a (-b / 2a) to find the vertex, we are given the x-value of the vertex.
We then plug this back into our equation to find Δy (Δy = ½ * g * t^2 + Voy * t + Yo).
We now know that the maximum height reached was .769 m.
(b) To find the ball’s range (or distance travelled horizontally), we use the formula x equals velocity in the x-direction times time plus initial position (X = Vx*t + Xo) to see how far it moved in the x-direction. We will use the Vx and t we found previously, and assume that the ball has a starting position of Om.
Therefore, the ball moved 11.475 m to the right.
(c) To increase the range, you could increase the velocity that it was kicked with (Vo) so it will travel faster and/or increase the angle at which it was kicked upward, which would keep the ball in the air for longer.
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