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?
The initial vertical velocity component is the same for both observers:
Vyo = 12.0 sin60 = 10.39 m/s
That is what determines how high it goes, H.
(1/2)M Vyo^2 = M g H
H = Vyo^2/(2g)= 5.51 m
To determine how high the professor sees the ball rise, we need to analyze the vertical motion of the ball. We will use the following kinematic equation to solve for the maximum height reached by the ball:
Δy = (v₀² sin²θ) / (2g)
Where:
Δy = maximum height reached by the ball
v₀ = initial velocity of the ball
θ = angle of projectile motion
g = acceleration due to gravity (approximately 9.8 m/s²)
Given:
v₀ = 12.0 m/s (because the train travels at a constant speed)
θ = 60.0° (because the student throws the ball at an initial angle of 60.0°)
g = 9.8 m/s²
Let's plug in the values into the equation:
Δy = (12.0² sin²60.0°) / (2 * 9.8)
To begin, let's calculate sin²60.0°:
sin²60.0° = (sin60.0°)²
= (0.866)²
= 0.749
Now let's substitute the values into the equation:
Δy = (12.0² * 0.749) / (2 * 9.8)
= 43.056 / 19.6
≈ 2.20 meters
Therefore, the professor sees the ball reach a maximum height of approximately 2.20 meters.