Final answer:
The position of the ball after 2.50 seconds is approximately 7.155 m, and its velocity is approximately 5.391 m/s. This is found by calculating the separate distances and velocities in the x and y directions, then vector summing them.
Step-by-step explanation:
The question involves a ball with an initial velocity along the y-axis and an acceleration in the x-direction. To find the position and velocity of the ball after 2.50 seconds, we can use the equations of motion for each axis separately since they are perpendicular to each other.
For the y-axis, the ball has no acceleration, so its velocity is constant at 1.30 m/s. For the x-axis, the ball starts from rest (initial velocity = 0) and accelerates at 2.10 m/s2.
For the y-axis, after 2.50 seconds, the ball's position (y) will be:
y = initial velocity * time
y = 1.30 m/s * 2.50 s = 3.25 m
For the x-axis, after 2.50 seconds, the ball's position (x) and velocity (vx) will be:
x = (1/2) * acceleration * time2
x = (1/2) * 2.10 m/s2 * (2.50 s)2 = 6.5625 m
vx = acceleration * time
vx = 2.10 m/s2 * 2.50 s = 5.25 m/s
The total position will be the vector sum of the positions in the x and y direction. Since these are at right angles to each other, we can use the Pythagorean theorem to find the total distance:
Total position = √(x2 + y2)
Total position = √(6.5625 m2 + 3.25 m2)
Total position = 7.155 m (approximately)
The total velocity of the ball will be the vector sum of the velocities in the x and y direction:
Total velocity = √(vx2 + vy2)
Total velocity = √(5.25 m/s2 + 1.30 m/s2)
Total velocity = 5.391 m/s (approximately)