Explain what an oblique motion is and how it can be decomposed into two distinct movements. Give examples of real situations that can be represented by an oblique trajectory. How can we calculate the time, displacement, and velocities in an oblique motion?

When a soccer player is very fast, a fun way to refer to this quality is to say that he is able to take a corner kick into the opponent's area and run himself to kick the ball before it touches the ground. Suppose a fictional player who is capable of doing this. If he takes the corner kick into the area providing the ball with an initial speed of $20 \mathrm{~m} / \mathrm{s}$, at an angle of $60^{\circ}$ with the horizontal, what distance does the player need to run, in a straight line, almost simultaneously with the corner kick, to kick the ball into the goal without letting it touch the ground? For simplification purposes, consider that the height of the goal kick is negligible, that $\operatorname{sen} 60^{\circ}=0.8, \cos 60^{\circ}=0.5$, and that the acceleration due to gravity is $10 \mathrm{~m} / \mathrm{s}^{2}$.

A stone of 0.5 kg is thrown upwards from a point located 8 m above the surface of the Earth, with a speed of 5 m/s. Disregarding dissipative forces of air resistance, regarding the movement of the stone, relate the left column with the right one. (I) Maximum height, in meters, that the stone can reach. (II) Speed, in m/s, immediately before hitting the ground. (III) Height, in meters, of the stone when the speed is equal to 3 m/s. (IV) Speed of the stone, in m/s, when it is at 4 m height.

A projectile is launched from a height H with an initial velocity V0 of 20 m/s, forming an angle of 30 degrees with the horizontal. Disregarding air resistance and considering gravity acceleration as g = 10 m/s^2, calculate the time it takes for the projectile to reach the ground and the horizontal range from the launch point to the impact point, assuming the ground is level. Consider the projectile's motion as a combination of uniform horizontal motion and uniformly accelerated vertical motion. Use sin(30 degrees) = 0.5 and cos(30 degrees) = 0.87 to facilitate calculations.

An object is fired obliquely with a speed of 10 m/s, initially forming an angle of 45 degrees with the horizontal. The acceleration due to gravity is equal to 10 m/s^2 and sin 45 degrees = cos 45 degrees = sqrt(2)/2. Considering x as the horizontal distance and y as the vertical distance traveled by the object, the difference x - y, in meters, at the highest point of the trajectory, is

To solve this question, we will follow these steps:

1. Decompose the initial velocity into horizontal and vertical components.
2. Calculate the time of ascent to the highest point of the trajectory.
3. Calculate the horizontal distance traveled to the highest point.
4. Calculate the maximum height reached (vertical distance).
5. Calculate the difference between the horizontal distance and the vertical distance at the highest point.

Step 1: Decomposition of the initial velocity
The initial velocity of 10 m/s is given at an angle of 45 degrees. Let's decompose this velocity into the horizontal component (Vx) and vertical component (Vy) using the trigonometric functions sine and cosine:

Vx = V * cos(45 degrees) = 10 m/s * (√2/2) = 5√2 m/s
Vy = V * sin(45 degrees) = 10 m/s * (√2/2) = 5√2 m/s

Step 2: Calculate the time of ascent
The vertical component of the velocity will be reduced to zero at the highest point of the trajectory due to gravity. We use the kinematic formula for uniformly accelerated motion:

Vy = Vy0 - g * t
0 = 5√2 m/s - 10 m/s^2 * t
t = (5√2 m/s) / (10 m/s^2)
t = √2 / 2 s

Step 3: Calculate the horizontal distance traveled
The horizontal distance traveled to the highest point is given by the horizontal component of the velocity multiplied by the time of ascent:

x = Vx * t
x = 5√2 m/s * √2 / 2 s
x = 5 m

Step 4: Calculate the maximum height
The maximum height is calculated using the vertical component of the velocity and the time of ascent:

y = Vy0 * t - (1/2) * g * t^2
y = 5√2 m/s * √2 / 2 s - (1/2) * 10 m/s^2 * (√2 / 2 s)^2
y = 5 m - (1/2) * 10 m/s^2 * (1/2)
y = 5 m - 5/4 m
y = 5 m - 1.25 m
y = 3.75 m

Step 5: Calculate the difference x - y
Now, we subtract the maximum height from the horizontal distance to find the difference at the highest point:

x - y = 5 m - 3.75 m
x - y = 1.25 m

The correct answer is 1.25 m, but this option is not available in the provided alternatives. Therefore, there seems to be an error in the question formulation or the presented alternatives. The calculated difference is 1.25 m, which would be closer to alternative (b) 1.0 m, but does not exactly match any of the options.

Question bank: Kinematics: Oblique Motion

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