A science student is riding on a flatcar of a train traveling along a strait, horizontal track at a consistent speed of 12 m/s. The student throws a ball into the air along a path he judges to make an initial angle of 65 degrees 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 he see the ball rise?
To determine how high the professor sees the ball rise, we need to consider the motion of the ball as observed by both the student on the flatcar and the professor on the ground. Let's break it down step by step:
Step 1: Analyze the motion of the ball as observed by the student on the flatcar.
The student sees the ball being thrown into the air at an angle of 65 degrees with the horizontal and in line with the track. Since there is no vertical acceleration (ignoring air resistance), the vertical motion can be treated independently from the horizontal motion.
Step 2: Separate the motion into horizontal and vertical components.
The initial velocity of the ball can be divided into its horizontal and vertical components. The horizontal component remains constant at 12 m/s throughout the motion, as there is no horizontal acceleration.
The vertical component of the initial velocity can be found using trigonometry. The vertical component is given by the equation:
v_y = v_initial * sin(theta)
where v_y is the vertical component of the initial velocity, v_initial is the initial velocity of the ball (which we can assume as 12 m/s in this case), and theta is the angle of projection (65 degrees in this case). Therefore,
v_y = 12 m/s * sin(65 degrees)
Step 3: Determine the time of flight.
The time it takes for the ball to reach its highest point in the air can be found by dividing the vertical displacement (rise) by the vertical component of velocity. Since the highest point is reached when the vertical component of velocity becomes zero, we can use the formula:
v_y = v_initial - g * t
where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time of flight. Setting v_y to zero and rearranging the equation, we get:
t = v_initial / g
Step 4: Calculate the maximum height of the ball.
To find the maximum height, we need to determine the vertical displacement of the ball during the time of flight. The formula for vertical displacement can be expressed as:
d = v_initial * t - (1/2) * g * t²
where d is the displacement (maximum height) and t is the time of flight. Substituting the values we found in the previous steps, we can calculate the maximum height.
By following these steps and performing the calculations with the given values, you can determine how high the professor sees the ball rise.
To calculate how high the professor sees the ball rise, we need to analyze the vertical component of the ball's motion. Since the professor is standing on the ground, he is observing the ball's vertical motion.
The initial velocity of the ball can be split into two components: the horizontal component and the vertical component. The initial angle of 65 degrees with the horizontal tells us that the initial vertical velocity (Vyi) of the ball is equal to the initial velocity (Vi) multiplied by the sine of the angle.
Vyi = Vi * sin(θ)
Vyi = 12 m/s * sin(65°)
Vyi = 12 m/s * 0.9063
Vyi ≈ 10.875 m/s
Now, we can find the time (t) it takes for the ball to reach its peak height by using the formula:
Vf = Vi + gt
Since the ball reaches its peak, the final velocity (Vf) will be zero, and the acceleration (g) due to gravity is approximately 9.8 m/s^2. Substituting the values:
0 = 10.875 m/s + (-9.8 m/s^2) * t
Solving for t:
10.875 m/s = 9.8 m/s^2 * t
t = 10.875 m/s / 9.8 m/s^2
t ≈ 1.11 s
Now, we can find the maximum height (h) the ball reaches using the formula:
h = Viy * t + 1/2 * g * t^2
Substituting the values:
h = 10.875 m/s * 1.11 s + 1/2 * 9.8 m/s^2 * (1.11 s)^2
h ≈ 12.0625 m + 6.12702 m
h ≈ 18.1895 m
Therefore, the professor sees the ball rise approximately 18.1895 meters.