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
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
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.