A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 18.0 m/s. The student throws a ball along a path that she judges to make an initial angle of 68.0° 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 straight up. How high does the ball rise?
To find the maximum height the ball reaches, we need to analyze the projectile motion of the ball.
Given:
- Initial velocity of the ball, v₀ = 18.0 m/s
- Angle of projection, θ = 68.0°
Step 1: Resolve the initial velocity into its vertical and horizontal components.
- The vertical component of the velocity, v_y, is given by v₀ * sin(θ).
- The horizontal component of the velocity, v_x, is given by v₀ * cos(θ).
Step 2: Calculate the time the ball takes to reach the maximum height.
- The time of flight for the projectile can be calculated using the formula t = 2 * v_y / g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
Step 3: Calculate the maximum height, h.
- The height reached by the ball, at its maximum point, is given by h = v_y² / (2 * g), where g is the acceleration due to gravity.
Let's calculate the maximum height step by step.
Step 1:
v_y = v₀ * sin(θ) = 18.0 m/s * sin(68.0°)
v_y ≈ 16.48 m/s
v_x = v₀ * cos(θ) = 18.0 m/s * cos(68.0°)
v_x ≈ 6.99 m/s
Step 2:
t = 2 * v_y / g = 2 * 16.48 m/s / 9.8 m/s²
t ≈ 3.36 s (rounded to two decimal places)
Step 3:
h = v_y² / (2 * g) = (16.48 m/s)² / (2 * 9.8 m/s²)
h ≈ 14.10 m (rounded to two decimal places)
Therefore, the ball rises to a height of approximately 14.10 meters.
To determine how high the ball rises, we can use projectile motion equations. Let's break down the problem step by step:
1. Identify the known values:
- Initial velocity of the ball (v₀): The ball is thrown horizontally along the track, so its horizontal component of velocity remains at 18.0 m/s.
- Initial angle with the horizontal (θ): 68.0°
- Acceleration due to gravity (g): Approximately 9.8 m/s²
2. Analyze the motion:
Since the initial velocity of the ball in the vertical direction is 0 (the ball initially rises straight up), we can use the equation: v = v₀ + gt, where v is the final vertical velocity of the ball, v₀ is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
3. Calculate the time it takes for the ball to reach its highest point:
At the highest point, the vertical velocity of the ball will be 0. Therefore, we can rearrange the equation v = v₀ + gt to solve for t: 0 = v₀ + gt.
Substituting the known values: 0 = 0 + 9.8t.
Solving for t: t = 0 s.
4. Determine the height of the ball at its highest point:
The height at the highest point can be found using the equation: h = v₀t + 0.5gt², where h is the height, v₀ is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.
Substituting the known values: h = 0 + 0.5 * 9.8 * (0)².
Simplifying: h = 0.
Therefore, the ball does not rise above its initial position. It rises straight up and falls back to the same height as where it was thrown.
steve where did you get 101m from
evidently the horizontal component of the ball's velocity is 18.0m/s toward the rear of the train.
18.0 = v*cos68.0°
v = 48m/s
The vertical component is thus 48*sin68.0° = 44.5 m/s
the height above the train is thus
h = 44.5t - 4.9t^2
= t(44.5 - 4.9t)
h is max at t=4.54
h(4.54) = 101m