A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 12.0 m/s. The student throws a ball into the air along a path that he judges to make an initial angle of 60.0° with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does she see the ball rise?
To determine the height to which the professor sees the ball rise, we need to break down the motion of the ball into horizontal and vertical components.
The horizontal component of the ball's motion is in line with the track, so it will not affect the vertical displacement. Therefore, the distance traveled in the horizontal direction by the time the ball reaches its maximum height is not relevant to finding the height.
The vertical component of the ball's motion is affected by gravity. In this case, the ball is initially thrown vertically but with an initial angle of 60.0°. We can focus on the vertical motion and use the equations of motion to find the maximum height.
Let's break down the initial velocity of the ball into its vertical and horizontal components:
Vertical component of initial velocity (Vy) = Velocity x sin(angle)
Horizontal component of initial velocity (Vx) = Velocity x cos(angle)
In this case:
Velocity = 12.0 m/s (given)
Angle = 60.0° (given)
Therefore, the vertical component of the initial velocity is:
Vy = 12.0 m/s x sin(60.0°)
Using the trigonometric value of sin(60.0°) = √3/2, we can calculate Vy:
Vy = 12.0 m/s x (√3/2)
Vy = 10.392 m/s
Now, we can use the equations of motion to find the maximum height (H) reached by the ball:
Vy^2 = Vo^2 + 2aH
Since the ball is at its maximum height, Vy = 0 m/s (ball stops moving upward momentarily, its velocity = 0), and acceleration (a) is -9.8 m/s^2 (due to gravity acting downwards):
0^2 = (10.392 m/s)^2 + 2 x -9.8 m/s^2 x H
Simplifying the equation:
0 = 108.0 m^2/s^2 - 19.6 m/s^2 x H
Solving for H, we find:
H = 108.0 m^2/s^2 / (19.6 m/s^2) = 5.51 m
Therefore, the professor sees the ball rise to a height of approximately 5.51 meters.