# Another motion in 2 dimensions ?

• 2slowtogofast
In summary, the ball thrown at 25.0 m/s and at an angle of 40 degrees above the horizontal hits the wall 18.8 m above the release point, with horizontal and vertical components of velocity at 19.7 m/s and 16.4 m/s respectively. The ball has also passed the highest point on its trajectory.
2slowtogofast
You throw a ball toward a wall at speed 25.0 m/s and at angle 40 degrees above the horizontal . The wall is distance d=22.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory?

as for A could you
could you treat it as s right triangle and since d = 22 m that could be the adjacent side. Then using that and the angle solve for the other two sides giving you the height?

a) The ball hits the wall 18.8 m above the release point. b) The horizontal component of its velocity as it hits the wall is 19.7 m/s. c) The vertical component of its velocity as it hits the wall is 16.4 m/s. d) Yes, it has passed the highest point on its trajectory.

Yes, that is a correct approach. Since we know the initial velocity (25.0 m/s) and the angle (40 degrees) at which the ball was thrown, we can use basic trigonometry to calculate the vertical component of its velocity. The horizontal component will remain constant at 25.0 m/s. So, for part (a), we can use the equation d = v0t + 1/2at^2 and solve for t, which will give us the time it takes for the ball to hit the wall. Then, using this time, we can use the equation vf = v0 + at to calculate the vertical component of the ball's velocity at the time of impact. This will give us the height at which the ball hits the wall. For parts (b) and (c), we can simply use basic trigonometry to calculate the horizontal and vertical components of the ball's velocity at the time of impact. As for part (d), we can determine if the ball has passed the highest point on its trajectory by comparing the vertical component of its velocity at the time of impact to its initial vertical velocity. If the vertical component at impact is less than the initial vertical velocity, then the ball has not yet reached its highest point.

## 1. What is "Another motion in 2 dimensions"?

Another motion in 2 dimensions refers to the movement of an object in both the x and y directions simultaneously. This is often seen in projectile motion, where an object is thrown or launched at an angle.

## 2. How is velocity calculated in 2 dimensions?

Velocity in 2 dimensions is calculated by finding the rate of change in both the x and y directions. This is done by dividing the change in position by the change in time in each direction separately, and then combining the two components using vector addition.

## 3. What is the difference between displacement and distance in 2 dimensions?

Displacement in 2 dimensions refers to the change in position of an object from its starting point to its ending point, taking into account both the x and y directions. Distance, on the other hand, refers to the total length of the path traveled by the object, regardless of direction.

## 4. Can an object have constant speed but changing velocity in 2 dimensions?

Yes, an object can have constant speed but changing velocity in 2 dimensions. This happens when the object is changing direction, even if its speed remains the same. Velocity is a vector quantity, meaning it takes into account both speed and direction.

## 5. How does air resistance affect motion in 2 dimensions?

Air resistance, also known as drag, can have a significant impact on motion in 2 dimensions. It can slow down the speed of an object and change its trajectory, making it deviate from its expected path. This is particularly important in projectile motion, where the shape and size of an object can greatly affect the amount of air resistance it experiences.

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