playing shortshop, you pick up a ground ball and throw it to second base. the ball is thrown horizontally with a speed of 22 m/s ditectly toward point a. when the ball reaches the second baseman 0.45 s later, it is caught at ponit b. (a) how far were you from the second baseman? (b) what is the distance of vertical drop ab?
To solve this problem, we can use the equations of motion and some basic geometry.
(a) To find the horizontal distance between you and the second baseman (point B), we need to determine the horizontal displacement (distance) traveled by the ball. We can use the formula:
horizontal displacement = horizontal velocity × time
Given:
Horizontal velocity (vx) = 22 m/s
Time (t) = 0.45 s
Substituting the values into the formula:
horizontal displacement = 22 m/s × 0.45 s = 9.9 m
Therefore, you were 9.9 meters away from the second baseman.
(b) To find the vertical distance of drop AB, we need to calculate the vertical displacement (height) traveled by the ball. Since the ball is thrown horizontally and there is no initial vertical velocity, the vertical motion is due to the force of gravity acting on the ball.
We can use the equation for vertical displacement:
vertical displacement = (1/2) × acceleration due to gravity × time²
The acceleration due to gravity is approximately 9.8 m/s².
Substituting the values into the equation:
vertical displacement = (1/2) × 9.8 m/s² × (0.45 s)² = 0.99 m
Therefore, the vertical distance of drop AB is approximately 0.99 meters.
To answer these questions, we can use the formulas of horizontal motion and vertical motion.
First, let's calculate the horizontal distance between you and the second baseman:
(a) The horizontal speed of the ball remains the same throughout its motion. So, the distance traveled horizontally is given by:
Distance = Speed x Time
Distance = 22 m/s x 0.45 s
Distance = 9.9 meters
Therefore, you were 9.9 meters away from the second baseman.
Now, let's calculate the vertical distance or drop between points A and B:
(b) We can use the formula for vertical motion:
Vertical Distance = (1/2) x g x t^2
Where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time it takes for the ball to reach point B (0.45s).
Vertical Distance = (1/2) x 9.8 m/s^2 x (0.45 s)^2
Vertical Distance ≈ 0.99 meters
Therefore, the vertical drop AB is approximately 0.99 meters.