A science student riding on a ﬂatcar of a train moving at a constant speed of a 9.70 m/s throws a ball toward the caboose along a path that the student judges as making an initial angle of 45.0◦ with the horizontal. The teacher, who is standing on the ground nearby,observes the ball rising vertically.

Question

A science student riding on a ﬂatcar of a train moving at a constant speed of a 9.70 m/s throws a ball toward the caboose along a path that the student judges as making an initial angle of 45.0◦ with the horizontal. The teacher, who is standing on the ground nearby,observes the ball rising vertically.

A.) How high does the ball rise? The acceleration of gravity is 9.81 m/s^2. Answer in units of m

Answer 1

Vo = 9.70m/s @ 45o.

Yo = 9.7*sin45 = 6.86 m/s. = Ver. comp.

h = (Y^2-Yo^2)/2g.
h = (0-47)/-19.6 = 2.4 m.

Answer 2

To determine how high the ball rises, we can use the kinematic equation for vertical motion:

h = (v₀² sin² θ) / (2g)

where:
h = height
v₀ = initial velocity
θ = angle with the horizontal
g = acceleration due to gravity

Given:
v₀ = 9.70 m/s
θ = 45.0°
g = 9.81 m/s²

First, convert the angle to radians:
θ_rad = 45.0° * (π / 180°) = 0.7854 rad

Now, substitute the values into the formula:
h = (9.70 m/s)² * sin²(0.7854 rad) / (2 * 9.81 m/s²)

h = (94.09 m²/s²) * (0.5) / (19.62 m/s²)
h = 4.623 m

Therefore, the ball rises to a height of 4.623 m.

Answer 3

To determine how high the ball rises, we need to analyze the projectile motion of the ball. We can break the initial velocity of the ball into horizontal and vertical components.

The horizontal component of the initial velocity remains constant throughout the motion because there are no horizontal forces acting on the ball. Therefore, the horizontal velocity remains constant at 9.70 m/s.

The vertical component of the initial velocity can be determined using the given angle of 45.0 degrees with the horizontal. We can use the trigonometric function sine to find the vertical component:

Vertical component of initial velocity (v₀y) = Initial velocity (v₀) × sin(θ)
v₀y = 9.70 m/s × sin(45.0 degrees)
v₀y = 9.70 m/s × 0.7071
v₀y ≈ 6.868 m/s

Now we can analyze the vertical motion of the ball. The ball follows a parabolic path due to the acceleration of gravity.
The maximum height (h) reached by the ball can be found using the following equation:

h = (v₀y^2) / (2 × g)

Substituting the values, we get:

h = (6.868 m/s)^2 / (2 × 9.81 m/s^2)
h ≈ 2.41 m

Therefore, the ball rises to a height of approximately 2.41 meters.